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HOW TO OPTIMIZE FOR THE PENTIUM PROCESSORS
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Copyright (c) 1996, 1997 by Agner Fog. Last modified 1997-05-03.
Contents:
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1. introduction
2. literature
3. debugging and verifying
4. memory model
5. alignment
6. cache
7. address generation interlock
8. pairing integer instructions
9. first time vs. repeated execution
10. imperfect pairing
11. splitting complex instructions into simpler ones
12. jumps and branches
13. prefixes
14. reducing code size
15. scheduling floating point code
16. loops
17. discussion of special instructions
18. using integer instructions to do floating point operations
19. using floating point instructions to do integer operations
20. list of integer instructions
21. list of floating point instructions
22. MMX instructions
23. testing code speed
24. Pentium Pro and Pentium II processors
25. considerations for other microprocessors
1. INTRODUCTION
===============
This manual describes in detail how to write optimized assembly language
code, with particular focus on the Pentium® microprocessor.
This manual is more comprehensive and exact than other sources of
information and contains many details not found anywhere else.
This information will enable you in many cases to calculate exactly how
many clock cycles a piece of code will take.
This manual covers the two versions of the Pentium: with and without MMX.
Differences between the two are mentioned where appropriate. The manual
will be updated later with more information on the Pentium with MMX. The
PentiumPro and PentiumII processors are very different, and will be
covered only briefly. I have no plans of doing research on the PentiumPro
and PentiumII processors.
The information in this manual is based on my own research, supplemented
with information I have received from various people. I want to thank
those who have sent me additional information for this manual.
Programming in assembly language is much more difficult than high level
language. Making bugs is very easy, and finding them is very difficult.
Now you have been warned! It is assumed that the reader is already
experienced in assembly programming. If not, then please read some books
on the subject and get some programming experience before you begin to do
complicated optimations.
The hardware design of the Pentium chip has many features which are
optimized specifically for some commonly used instructions or instruction
combinations, rather than using general optimation methods. Consequently,
the rules for optimizing software for this design are quite complicated
and have many exceptions, but the possible gain in performance may be
substantial.
Before you start to convert your code to assembly, make sure that your
algorithm is optimal. Often you can improve a piece of code much more by
improving the algorithm than by converting it to assembler code.
Next, you have to identify the critical part of your program. Often more
than 99% of the CPU time is used in the innermost loop of a program. In
this case you should optimize only that loop and leave everything else
in high level language. Some assembly programmers waste a lot of energy
optimizing the wrong parts of their programs, the only significant effect
of their effort being that the programs become more difficult to debug
and maintain!
If it is not obvious where the critical parts of your program are then
you may use a profiler to find them. If it turns out that the bottleneck
is disk access, then you may modify your program to reduce the number of
disk accesses rather than turning to assembly programming. If the
bottleneck is graphics output then you may look for a way of reducing
the number of calls to graphic procedures.
Some high level language compilers offer relatively good optimation for
the Pentium, and further optimation by hand may not be worth the effort
except for the most critical pieces of code - or for the sport of it.
Please don't send your programming questions to me. I am not gonna do your
homework for you.
Good luck with your hunt for nanoseconds!
PS. I have deliberately misspelled 'optimization' because I think
'optimation' is more optimal to pronounce.
2. LITERATURE
=============
A lot of useful literature can be downloaded for free from Intel's www
site or acquired in print or on CD-ROM.
I will not give the URL's here because they change very often.
You can find the documents you need by using the search facilities at:
http://www.intel.com/sites/developer/search.htm or follow the links from
http://announce.com/agner/assem
Some documents are in .PDF format. If you don't have software for viewing
or printing .PDF files, then you may download the Acrobat file reader from
www.adobe.com
The most useful document is Intel's application note:
'AP-526 Optimizations for Intel's 32-bit Processors'
Beginners may use a funny tutorial named "Optimizing Applications for the
Pentium Processor"
Optimations for the PentiumPro and PentiumII processors are described in
'AP-526 Optimizations for Intel's 32-bit Processors' and in the tutorial
'Optimizing for the Pentium® Pro Processor Family'.
The use of MMX instructions for optimizing specific applications are
described in several application notes. The MMX instruction set is
described in various manuals and tutorials.
A lot of other sources than Intel also have useful information. These
sources are listed in the FAQ for the newsgroup comp.lang.asm.x86.
The shareware editor ASMEDIT has an online help which covers all
instruction codes etc. It is available from:
http://www.inf.tu-dresden.de/~ok3/asmedit.html
For other internet ressources follow the links from
http://announce.com/agner/assem/
3. DEBUGGING AND VERIFYING
==========================
Debugging assembly code can be quite hard and frustrating, as you probably
already have discovered. I would recommend that you start with writing the
piece of code you want to optimize as a subroutine in a high level language.
Next, write a test program that will test your subroutine thoroughly. Make
sure the test program goes into all branches and special cases.
When your high level language subroutine works with your test program, then
you are ready to translate the code to assembly language (some high level
compilers will do this for you), and test it on the test program.
Then you can start to optimize. Each time you have made a modification you
should run it on the test program to see if it works correctly.
Number all your versions and save them so that you can go back and test
them again in case you discover an error which the test program didn't
catch (such as writing to a wrong address).
Test the speed of the most critical part of your program with the method
described in chapter 23. If the code is significantly slower than expected,
then the most probable reasons are: cache misses (chapter 6), misaligned
operands (chapter 5), first time penalty (chapter 9), and branch
mispredictions (chapter 12).
4. MEMORY MODEL
===============
The Pentium is designed primarily for 32 bit code, and it's performance is
inferior on 16 bit code. Segmenting your code and data also degrades
performance significantly, so you should generally prefer 32 bit flat mode,
and an operating system which supports this mode (e.g. Windows 95, OS/2, or
a 32 bit DOS extender). The code examples shown in this manual assume a 32
bit flat memory model, unless otherwise specified.
5. ALIGNMENT
============
All data in RAM should be aligned to addresses divisible by 2, 4, or 8
according to this scheme:
operand size alignment
---------------------------
1 (byte) 1
2 (word) 2 (or address MOD 4 >< 3. other proc. require align by 2)
4 (dword) 4
6 (fword) 4 (Pentium Pro requires alignment by 8)
8 (qword) 8
10 (tbyte) 8 (Pentium Pro requires alignment by 16)
Misaligned data will take at least 3 clock cycles extra to access.
Aligning code is not necessary when you run on the Pentium, but for optimal
performance on other processors you may align subroutine entries and loop
entries by 16.
6. CACHE
========
The Pentium without MMX has 8 kb of on-chip cache (level one cache) for
code, and 8 kb for data. The Pentium with MMX has 16 kb for code and 16 kb
for data. Data in the level 1 cache can be read or written to in just one
clock cycle, whereas a cache miss may cost many clock cycles. It is
therefore important that you understand how the cache works in order to use
it most efficiently.
The data cache in the Pentium without MMX consists of 256 lines of 32 bytes
each. Each time you read a data item which is not cached, the processor will
read an entire cache line from memory. The cache lines are always aligned to
a physical address divisible by 32. When you have read a byte at an address
divisible by 32, then the next 31 bytes can be read or written to at almost
no extra cost. You can take advantage of this by arranging data items which
are used near each other together into aligned blocks of 32 bytes of memory.
If, for example, you have a loop which accesses two arrays, then you may
interleave the two arrays into one array of structures, so that data which are
used together are also stored together.
If the size of an array or other data structure is a multiple of 32 bytes,
then you should preferably align it by 32.
A cache line can not be assigned to an arbitrary memory address. Each
cache line has a 7-bit set-value which must match bits 5 through 11 of the
physical RAM address (bit 0-4 define the 32 bytes within a cache line).
The Pentium without MMX has two cache lines for each of the 128 set-values,
so there are two possible cache lines to assign to any RAM address. The
MMX version has four.
The consequence of this is that the cache can hold no more than two or
four different data blocks which have the same value in bits 5-11 of the
address. You can determine if two addresses have the same set-value by
the following method: Strip off the lower 5 bits of each address to get
a value divisible by 32. If the difference between the two truncated
addresses is a multiple of 4096 (=1000H), then the addresses have the
same set-value.
Let me illustrate this by the following piece of code, where ESI holds an
address divisible by 32:
AGAIN: MOV EAX, [ESI]
MOV EBX, [ESI + 13*4096 + 4]
MOV ECX, [ESI + 20*4096 + 28]
DEC EDX
JNZ AGAIN
The three addresses used here all have the same set-value because the
differences between the truncated addresses are multipla of 4096. This
loop will perform very poorly on the Pentium without MMX. At the time
you read ECX there is no free cache line with the proper set-value so
the processor takes the least recently used of the two possible cache
lines, that is the one which was used for EAX, and fills it with the
data from [ESI + 20*4096] to [ESI + 20*4096 + 31] and reads ECX. Next,
when reading EAX, you find that the cache line that held the value for
EAX has now been discarded, so you take the least recently used line,
which is the one holding the EBX value, and so on.. You have nothing
but cache misses and the loop takes something like 60 clock cycles.
If the third line is changed to:
MOV ECX, [ESI + 20*4096 + 32]
then we have crossed a 32 byte boundary, so that we do not have the same
set-value as in the first two lines, and there will be no problem assigning
a cache line to each of the three addresses. The loop now takes only 3 clock
cycles (except for the first time) - a very considerable improvement!
It may be very difficult to determine if your data addresses have the
same set-values, especially if they are scattered around in different
segments. The best thing you can do to avoid problems of this kind is
to keep all data used in the critical part or your program within one
contiguous block not bigger than the cache, or two contiguous blocks
no bigger than half that size (for example one block for static data
and another block for data on the stack). This will make sure that
your cache lines are used optimally.
If the critical part of your code accesses big data structures or random
data addresses, then you may want to keep all frequently used variables
(counters, pointers, control variables, etc.) within a single contiguous
block of max 4 kbytes so that you have a complete set of cache lines free
for accessing random data. Since you probably need stack space anyway for
subroutine parameters and return addresses, the best thing is to copy all
frequently used static data to dynamic variables on the stack, and copy
them back again outside the critical loop if they have been changed.
Reading a data item which is not in the level one cache causes an entire
cache line to be filled from the level two cache, which takes approximately
200 ns (that is 20 clocks on a 100 MHz system or 40 clocks on a 200 MHz
system), but the bytes you ask for first are available already after 50-100
ns. If the data item is not in the level two cache either, then you will
get a delay of something like 200-300 ns. This delay will be somewhat longer
if you cross a DRAM page boundary. (The size of a DRAM page is 1 kb for
4 and 8 MB 72 pin RAM modules, and 2 kb for 16 and 32 MB modules).
When you write to an address which is not in the level 1 cache, then the
value will go right through to the level 2 cache or to the RAM (depending on
how the level 2 cache is set up). This takes approximately 100 ns. If you
write eight or more times to the same 32 byte block of memory without also
reading from it, and the block is not in the level one cache, then it may
be advantageous to make a dummy read from the block first to load it into a
cache line. All subsequent writes to the same block will then go to the
cache instead, which takes only one clock cycle. (This is not needed on a
PentiumPro or PentiumII, which always load a cache line on write misses).
There is sometimes a small penalty for writing repeatedly to the same
address without reading in between.
Another way to reduce the number of write misses is to write 8 bytes at a
time using FILD / FISTP with qword operands rather than writing 4 bytes at
a time using integer registers. The extra cost of the slow FILD and FISTP
instructions is compensated for by the fact that you only have to do half as
many writes. However, this method is only advantageous on a Pentium, and
only if the destination is not in the level 1 cache. The method is explained
in section 19 below. If you have the MMX version you may of course use MMX
instructions rather than floating point instructions to move 8 bytes at a
time.
Temporary data may conveniently be stored on the stack because the stack
area is very likely to be in the cache. You should be aware, however,
that if you store QWORD data on a DWORD size stack, or DWORD data on a
WORD size stack, then you have a potential alignment problem.
If the life ranges of two data structures do not overlap, then they may
use the same RAM area to increase cache efficiency. This is consistent
with the common practice of allocating space for temporary variables on
the stack.
Storing temporary data in registers is of course even more efficient.
Since registers is a scarce ressource you may want to use [ESP] rather
than [EBP] for addressing data on the stack, in order to free EBP for
other purposes. Just don't forget that the value of ESP changes every
time you do a PUSH or POP. (You cannot use ESP under 16-bit Windows
because the timer interrupt will modify the high word of ESP at
unpredictable places in your code.)
There is a separate cache for code, which is similar to the data cache.
The size of the code cache is 8 kb on the versions without MMX and 16 kb
on the MMX versions. It is important that the critical part of your code
(the innermost loops) fit in the code cache. Frequently used pieces of
code or routines which are used together should preferable be stored near
each other. Seldom used branches or procedures should be put away in the
bottom of your code or somewhere else.
7. ADDRESS GENERATION INTERLOCK (AGI)
=====================================
It takes one clock cycle to calculate the address needed by an instruction
which accesses memory. Normally, this calculation is done at a separate
stage in the pipeline while the preceding instruction or instruction pair is
executing. But if the address depends on the result of an instruction
executing in the preceding clock cycle, then you have to wait an extra clock
cycle for the address to be calculated. This is called an AGI stall.
Example:
ADD EBX,4 / MOV EAX,[EBX] ; AGI stall
The stall in this example can be removed by putting some other instructions
in between ADD EBX,4 and MOV EAX,[EBX] or by rewriting the code to:
MOV EAX,[EBX+4] / ADD EBX,4
You can also get an AGI stall with instructions which use ESP implicitly for
addressing, such as PUSH, POP, CALL, and RET, if ESP has been changed in the
preceding clock cycle by instructions such as MOV, ADD, or SUB. The Pentium
has special circuitry to predict the value of ESP after a stack operation so
that you do not get an AGI delay after changing ESP with PUSH, POP, or CALL.
You can get an AGI stall after RET only if it has an immediate operand to
add to ESP.
Examples:
ADD ESP,4 / POP ESI ; AGI stall
POP EAX / POP ESI ; no stall, pair
MOV ESP,EBP / RET ; AGI stall
CALL L1 / L1: MOV EAX,[ESP+8] ; no stall
RET / POP EAX ; no stall
RET 8 / POP EAX ; AGI stall
The LEA instruction is also subject to an AGI stall if it uses a base or
index register which has been changed in the preceding clock cycle. Example:
INC ESI / LEA EAX,[EBX+4*ESI] ; AGI stall
8. PAIRING INTEGER INSTRUCTIONS
===============================
The Pentium has two pipelines for executing instructions, called the U-pipe
and the V-pipe. Under certain conditions it is possible to execute two
instructions simultaneously, one in the U-pipe and one in the V-pipe. This
can almost double the speed. It is therefore advantageous to reorder your
instructions to make them pair.
The following instructions are pairable in either pipe:
MOV register, memory, or immediate into register or memory
PUSH register or immediate
POP register
LEA, NOP
INC, DEC, ADD, SUB, CMP, AND, OR, XOR,
and some forms of TEST (see section 17.3)
The following instructions are pairable in the U-pipe only:
ADC, SBB
SHR, SAR, SHL, SAL with immediate count
ROR, ROL, RCR, RCL with an immediate count of 1
The following instructions can execute in either pipe but are only pairable
when in the V-pipe: near call, short and near jump, short and near
conditional jump.
All other integer instructions can execute in the U-pipe only, and are not
pairable.
Two consecutive instructions will pair when the following conditions are met:
8.1 The first instruction is pairable in the U-pipe and the second
instruction is pairable in the V-pipe.
8.2 The second instruction cannot read or write a register which the first
instruction writes to. Examples:
MOV EAX, EBX / MOV ECX, EAX ; read after write, do not pair
MOV EAX, 1 / MOV EAX, 2 ; write after write, do not pair
MOV EBX, EAX / MOV EAX, 2 ; write after read, pair OK
MOV EBX, EAX / MOV ECX, EAX ; read after read, pair OK
MOV EBX, EAX / INC EAX ; read and write after read, pair OK
8.3 In rule 8.2 partial registers are treated as full registers. Example:
MOV AL, BL / MOV AH, 0 ; writes to different parts of the same
; register, do not pair
8.4 Two instructions which both write to parts of the flags register can
pair despite rule 8.2 and 8.3. Example:
SHR EAX,4 / INC EBX ; pair OK
8.5 An instruction which writes to the flags can pair with a conditional
jump despite rule 8.2. Example:
CMP EAX, 2 / JA LabelBigger ; pair OK
8.6 The following instruction combinations can pair despite the fact that
they both modify the stack pointer:
PUSH + PUSH, PUSH + CALL, POP + POP
8.7 There are restrictions on the pairing of instructions with prefix.
There are several types of prefixes:
- instructions addressing a non-default segment have a segment prefix.
- instructions using 16 bit data in 32 bit mode, or 32 bit data in 16 bit
mode have an operand size prefix.
- instructions using 32 bit base or index registers in 16 bit mode have
an address size prefix.
- repeated string instructions have a repeat prefix.
- locked instructions have a LOCK prefix.
- many instructions which were not implemented on the 8086 processor
have a two byte opcode where the first byte is 0FH. The 0FH byte
behaves as a prefix on the Pentium without MMX, but not on the Pentium
with MMX. The most common instructions with 0FH prefix are: MOVZX,
MOVSX, PUSH FS, POP FS, PUSH GS, POP GS, LFS, LGS, LSS, SETcc, BT,
BTC, BTR, BTS, BSF, BSR, SHLD, SHRD, and IMUL with two operands and no
immediate operand.
On the Pentium without MMX, a prefixed instruction can only execute in
the U-pipe, except for conditional near jumps.
On the Pentium with MMX, instructions with operand size, address
size, or 0FH prefix can execute in either pipe, whereas instructions with
segment, repeat, or lock prefix can only execute in the U-pipe.
8.8 An instruction which has both a displacement and immediate data is not
pairable on the Pentium without MMX and only pairable in the U-pipe on
Pentium with MMX:
MOV DWORD PTR DS:[1000], 0 ; not pairable or only in U-pipe
CMP BYTE PTR [EBX+8], 1 ; not pairable or only in U-pipe
CMP BYTE PTR [EBX], 1 ; pairable
CMP BYTE PTR [EBX+8], AL ; pairable
(Another problem with instructions which have both a displacement and
immediate data on the Pentium with MMX is that such instructions may
be longer than 7 bytes, which means that only one instruction can be
decoded per clock cycle, as explained in section 13.)
8.9 Both instructions must be preloaded and decoded. This is explained in
section 9.
8.10 There are special pairing rules for MMX instructions:
- MMX shift, pack or unpack instructions can execute in either pipe
but cannot pair with other MMX shift, pack or unpack instructions.
- MMX multiply instructions can execute in either pipe but cannot pair
with other MMX multiply instructions. They take 3 clock cycles and
the last 2 clock cycles can overlap with subsequent instructions
in the same way as floating point instructions can (see chapter 15).
- an MMX instruction which accesses memory or integer registers can
execute only in the U-pipe and cannot pair with a non-MMX instruction.
9. FIRST TIME VERSUS REPEATED EXECUTION
=======================================
A piece of code usually takes much more time the first time it is executed
than when it is repeated. The reasons are the following:
9.1 Loading the code from RAM into the cache takes longer time than
executing it.
9.2 In some cases decoding the code is the bottleneck. If it takes one clock
cycle to determine the length of an instruction, then it is not possible
to decode two instructions per clock cycle, because the processor
doesn't know where the second instruction begins. The Pentium solves
this problem by remembering the length of any instruction which has
remained in the cache since last time it was executed. As a consequence
of this, a set of instructions will not pair the first time they are
executed, unless the first of the two instructions is only one byte
long.
9.3 Jump instructions will not be in the branch target buffer the first
time they execute, and therefore are less likely to be predicted
correctly. See chapter 12.
9.4 Any data accessed by the code has to be loaded into the cache, which may
take much more time than executing the instructions. When the code is
repeated, then the data are more likely to be in the cache.
For these four reasons, a piece of code inside a loop will generally take
much more time the first time it executes than the subsequent times.
If you have a big loop which doesn't fit into the code cache then you will
get the penalty of 9.1 and 9.2 all the time because it doesn't run from the
cache. You should therefore try to reorganize the loop to make it fit into
the cache.
If you have many jumps and branches inside a loop, then you may get the
penalty in 9.3 all the time.
Likewise, if a loop repeatedly accesses a data structure too big for
the data cache, then you will get the penalty of data cache misses all the
time.
10. IMPERFECT PAIRING
=====================
There are situations where the two instructions in a pair will not execute
simultaneously, or only partially overlap in time. They should still be
considered a pair, though, because the first instruction executes in the
U-pipe, and the second in the V-pipe. No subsequent instruction can start
to execute before both instructions in the imperfect pair have finished.
Imperfect pairing will happen in the following cases:
10.1 If the second instructions suffers an AGI stall (see chapter 7).
10.2 Two instructions cannot access the same dword of memory simultaneously.
The following examples assume that ESI is divisible by 4:
MOV AL, [ESI] / MOV BL, [ESI+1]
The two operands are within the same dword, so they cannot execute
simultaneously. The pair takes 2 clock cycles.
MOV AL, [ESI+3] / MOV BL, [ESI+4]
Here the two operands are on each side of a dword boundary, so they
pair perfectly, and take only one clock cycle.
10.3 Rule 10.2 is extended to the case where bit 2-4 is the same in the two
addresses (cache bank conflict). For dword addresses this means that
the difference between the two addresses should not be divisible by 32.
Examples:
MOV [ESI], EAX / MOV [ESI+32000], EBX ; imperfect pairing
MOV [ESI], EAX / MOV [ESI+32004], EBX ; perfect pairing
Pairable instructions which do not access memory take one clock cycle to
execute, except for mispredicted jumps. MOV instructions to or from memory
also take only one clock cycle if the data area is in the cache and properly
aligned. There is no speed penalty for using complex addressing modes such
as scaled index registers.
A pairable instruction which reads from memory, does some calculation, and
stores the result in a register or flags, takes 2 clock cycles.
(read/modify instructions).
A pairable instruction which reads from memory, does some calculation, and
writes the result back to the memory, takes 3 clock cycles.
(read/modify/write instructions).
10.4 If a read/modify/write instruction is paired with a read/modify or
read/modify/write instruction, then they will pair imperfectly.
The number of clock cycles used is given in the following table:
| First instruction
| MOV or read/ read/modify/
Second instruction | register only modify write
----------------------|----------------------------------------------
MOV or register only | 1 2 3
read/modify | 2 2 4
read/modify/write | 3 3 5
----------------------|-----------------------------------------------
Example:
ADD [mem1], EAX / ADD EBX, [mem2] ; 4 clock cycles
ADD EBX, [mem2] / ADD [mem1], EAX ; 3 clock cycles
10.5 When two paired instructions both take extra time due to cache misses,
misalignment, or jump misprediction, then the pair will take more time
than each instruction, but less than the sum of the two.
In order to avoid imperfect pairing you have to know which instructions go
into the U-pipe, and which to the V-pipe. You can find out this by looking
backwards in your code and search for instructions which are unpairable,
pairable only in one of the pipes, or cannot pair due to one of the rules
in chapter 8.
Imperfect pairing can often be avoided by reordering your instructions.
Example:
L1: MOV EAX,[ESI]
MOV EBX,[ESI]
INC ECX
Here the two MOV instructions form an imperfect pair because they both
access the same memory location, and the sequence takes 3 clock cycles.
You can improve the code by reordering the instructions so that INC ECX
pairs with one of the MOV instructions.
L2: MOV EAX,OFFSET A
XOR EBX,EBX
INC EBX
MOV ECX,[EAX]
JMP L1
The pair INC EBX / MOV ECX,[EAX] is imperfect because the latter
instruction has an AGI stall. The sequence takes 4 clocks. If you insert
a NOP or any other instruction so that MOV ECX,[EAX] pairs with JMP L1
instead, then the sequence takes only 3 clocks.
The next example is in 16 bit mode, assuming that SP is divisible by 4:
L3: PUSH AX
PUSH BX
PUSH CX
PUSH DX
CALL FUNC
Here the PUSH instructions form two imperfect pairs, because both operands
in each pair go into the same dword of memory. PUSH BX could possibly pair
perfectly with PUSH CX (because they go on each side of a dword boundary)
but it doesn't because it has already been paired with PUSH AX. The sequence
therefore takes 5 clocks. If you insert a NOP or any other instruction so
that PUSH BX pairs with PUSH CX, and PUSH DX with CALL FUNC, then the
sequence will take only 3 clocks. Another way to solve the problem is to
make sure that SP is not divisible by 4. Knowing whether SP is divisible by
4 or not in 16 bit mode can be difficult, so the best way to avoid this
problem is to use 32 bit mode.
11. SPLITTING COMPLEX INSTRUCTIONS INTO SIMPLER ONES
====================================================
You may split up read/modify and read/modify/write instructions to improve
pairing. Example:
ADD [mem1],EAX / ADD [mem2],EBX ; 5 clock cycles
This code may be split up into a sequence which takes only 3 clock cycles:
MOV ECX,[mem1] / MOV EDX,[mem2] / ADD ECX,EAX / ADD EDX,EBX
MOV [mem1],ECX / MOV [mem2],EDX
Likewise you may split up non-pairable instructions into pairable instructions:
PUSH [mem1] / PUSH [mem2] ; non-pairable
Split up into:
MOV EAX,[mem1] / MOV EBX,[mem2] / PUSH EAX / PUSH EBX ; everything pairs
Other examples of non-pairable instructions which may be split up into simpler
pairable instructions:
CDQ split into: MOV EDX,EAX / SAR EDX,31
NOT EAX change to XOR EAX,-1
NEG EAX split into XOR EAX,-1 / INC EAX
MOVZX EAX,BYTE PTR [mem] split into XOR EAX,EAX / MOV AL,BYTE PTR [mem]
JECXZ split into TEST ECX,ECX / JZ
LOOP split into DEC ECX / JNZ
XLAT change to MOV AL,[EBX+EAX]
If splitting instructions doesn't improve speed, then you may keep the
complex or nonpairable instructions in order to reduce code size.
Splitting instructions is not needed on the Pentium Pro and Pentium II.
12. JUMPS AND BRANCHES
======================
Note: This chapter describes in detail the branch prediction mechanism of
the Pentium without MMX. This information is based on research done by me
and Karki Jitendra Bahadur. Information found in Intel documents and
elsewhere on this subject is directly misleading, and following the
advises given is such documents is likely to lead to sub-optimal code.
The branch prediction mechanism on the MMX version is very different.
I don't have complete information on the MMX version yet.
In the following, I will use the term 'jump instruction' for any
instruction which can change the instruction pointer, including
conditional and unconditional, direct and indirect, near and far, jumps,
calls, and returns.
The Pentium attempts to predict where a jump will go to, and whether a
conditional jump will be taken or fall through. If the prediction is correct,
then it can save time by loading the subsequent instructions into the
pipeline before the jump is executed. If the prediction turns out to be
wrong, then the pipeline has to be flushed, which will cost a penalty of 3
to 5 clock cycles. (4 for a conditional jump in the V-pipe, 3 in all other
cases for the Pentium without MMX. The MMX version has penalties of 5 resp.
4 clocks.)
When optimizing code, it is important to minimize the number of misprediction
penalties. This requires a good understanding of how the jump prediction
works. I am giving a very detailed description of this topic here, because
this information is not published anywhere else.
Note that the following description applies only to the Pentium without
MMX. Branch prediction in later processors will only be covered briefly.
Branch prediction
-----------------
The Pentium uses a 'branch target buffer' (BTB), which can save information
for up to 256 jump instructions. The BTB is organized like a 4-way set-
associative cache with 64 entries per way. This means that the BTB can hold
no more than 4 entries with the same set value. Unlike the data cache, the
BTB uses a pseudo random replacement algorithm, which means that a new entry
will not necessarily displace the least recently used entry of the same set-
value. How the set-value is calculated will be explained later. Each BTB
entry stores the address of the jump target and a prediction state, which
can have four different values:
state 0: "strongly not taken"
state 1: "weakly not taken"
state 2: "weakly taken"
state 3: "strongly taken"
A branch instruction is predicted to jump when in state 2 or 3, and to
fall through when in state 0 or 1. Some documents do - not quite correctly -
describe the state transition so that the state is incremented when the
branch is taken, and decremented when it falls through (no wrap around).
This would provide quite a good prediction, because a branch instruction
would have to deviate twice from what it does most of the time, before the
prediction changes.
However, this mechanism has been compromised by the fact that state
0 also means 'unused BTB entry'. So a BTB entry in state 0 is the same as
no BTB entry. This makes sense, because a branch instruction is predicted
to fall through if it has no BTB entry. This improves the utilization of
the BTB, because a branch instruction which is seldom taken will most of
the time not take up any BTB entry.
Now, if a jumping instruction has no BTB entry, then a new BTB entry will
be generated, and this new entry will always be set to state 3. This means
that it is impossible to go from state 0 to state 1 (except for a very
special case discussed later). From state 0 you can only go to state 3, if
the branch is taken. If the branch falls through, then it will stay out of
the BTB.
This is a serious design flaw. By always setting new entries to state
3, the designers apparently have given priority to minimizing the first
time penalty for unconditional jumps and branches often taken, rather than
improving the prediction of conditional jumps in tight innermost loops,
which is almost the only place where speed really matters. The consequence
of this flaw is, that a branch instruction which falls through most of the
time will have up to three times as many mispredictions as a branch
instruction which is taken most of the time.
You may take this asymmetry into account by organizing your branches so
that they are taken more often than not.
Consider for example this if-then-else construction:
TEST EAX,EAX
JZ A
<branch 1>
JMP E
A: <branch 2>
E:
If branch 1 is executed more often than branch 2, and branch 2 is seldom
executed twice in succession, then you can reduce the number of branch
mispredictions by up to a factor 3 by swapping the two branches so that
the branch instruction will jump more often than fall through:
TEST EAX,EAX
JNZ A
<branch 2>
JMP E
A: <branch 1>
E:
(This is _contrary_ to the recommendations in Intel's tutorial. They don't
seem to recognize the asymmetry in branch prediction. But it is easy to
prove that the latter example is superior).
There may be reasons to put the most often executed branch first, however:
1. Putting seldom executed branches away in the bottom of your code can
improve code cache utilization.
2. A branch instruction seldom taken will stay out of the BTB most of the
time, possibly improving BTB utilization.
3. The branch instruction will be predicted as not taken if it has been
flushed out of the BTB by other branch instructions.
4. The asymmetry in branch prediction only exists on the Pentium without MMX (?)
These considerations have little weight, however, for small critical loops,
so I would still recommend organizing branches with a skewed distribution
so that the branch instruction is taken more often than not, unless branch
2 is executed so seldom, that misprediction doesn't matter.
Likewise, you should preferably organize loops with the testing branch at
the bottom, as in this example:
MOV ECX, [N]
L: MOV [EDI],EAX
ADD EDI,4
DEC ECX
JNZ L
If N is high, then the JNZ instruction here will be taken more often than
not, and never fall through twice in succession.
BTB is looking ahead
--------------------
The BTB mechanism is counting instruction pairs, rather than single
instructions, so you have to know how instructions are pairing in order
to analyze where a BTB entry is stored. The BTB entry for any jump
instruction is attached to the address of the U-pipe instruction in the
preceding instruction pair. (An unpaired instruction counts as one pair).
Example:
SHR EAX,1
MOV EBX,[ESI]
CMP EAX,EBX
JB L
Here SHR pairs with MOV, and CMP pairs with JB. The BTB entry for
JB L is thus attached to the address of the SHR EAX,1 instruction. When
this BTB entry is met, and if it is in state 2 or 3, then the Pentium will
read the target address from the BTB entry, and load the instructions
following L into the pipeline. This happens before the branch instruction
has been decoded, so the Pentium relies solely on the information in the
BTB when doing this.
You may remember, that instructions are seldom pairing the first time they
are executed (see paragraph 9.2). If the instructions above are not
pairing, then the BTB entry should be attached to the address of the CMP
instruction, and this entry would be wrong on the next execution, when
instructions are pairing. However, in most cases the Pentium is smart enough
to not make a BTB entry when there is an unused pairing opportunity, so you
don't get a BTB entry until the second execution, and hence you won't get a
prediction until the third execution. (In the rare case, where every second
instruction is a single-byte instruction, you may get a BTB entry on the
first execution which becomes invalid in the second execution, but since
the instruction it is attached to will then go to the V-pipe, it is ignored
and gives no penalty. A BTB entry is only read if it is attached to the
address of a U-pipe instruction).
A BTB entry is identified by its set-value which is equal to bits 0-5
of the address it is attached to. Bits 6-31 are then stored in the BTB as
a tag. Addresses which are spaced a multiple of 64 bytes apart will have
the same set-value. You can have no more than four BTB entries with the
same set-value. If you want to check whether your jump instructions contend
for the same BTB entries, then you have to compare bits 0-5 of the addresses
of the U-pipe instructions in the preceding instruction pairs. This is very
tedious, and I have never heard of anybody doing so. There are no tools
available to do this job for you, as far as I know.
Consecutive branches
--------------------
When a jump is mispredicted, then the pipeline gets flushed. If the next
instruction pair executed also contains a jump instruction, then the Pentium
won't load its target because it cannot load a new target while the
pipeline is being flushed. The result is that the second jump instruction
is predicted to fall through regardless of the state of its BTB entry.
Therefore, if the second jump is also taken, then you will get another
penalty. The state of the BTB entry for the second jump instruction does get
correctly updated, though. If you have a long chain of taken jump
instructions, and the first jump in the chain is mispredicted, then the
pipeline will get flushed all the time, and you will get nothing but
mispredictions until you meet an instruction pair that does not jump. The
most extreme case of this is a loop which jumps to itself: It will get a
misprediction penalty for each iteration.
This is not the only problem with consecutive jumps. Another problem is that
you can have a branch instruction between a BTB entry and the jump
instruction it belongs to. If the first branch instruction jumps to
somewhere else, then strange things may happen. Consider this example:
SHR EAX,1
MOV EBX,[ESI]
CMP EAX,EBX
JB L1
JMP L2
L1: MOV EAX,EBX
INC EBX
When JB L1 falls through, then you will get a BTB entry for JMP L2
attached to the address of CMP EAX,EBX. But what will happen when
JB L1 later is taken? At the time when the BTB entry for JMP L2 is
read, the processor doesn't know that the next instruction pair does not
contain a jump instruction, so it will actually predict the instruction
pair MOV EAX,EBX / INC EBX to jump to L2. The penalty for predicting
non-jump instructions to jump is 3 clock cycles. The BTB entry for JMP L2
will get its state decremented, because it is applied to something which
doesn't jump. If we keep going to L1, then the BTB entry for JMP L2
will be decremented to state 1 and 0, so that the problem will disappear
until next time JMP L2 is executed.
The penalty for predicting the non-jumping instructions to jump only occurs
when the jump to L1 is predicted. In the case that JB L1 is mispredictedly
jumping, then the pipeline gets flushed and we won't get the false L2 target
loaded, so in this case we will not see the penalty of predicting the
non-jumping instructions to jump, but we do get the BTB entry for JMP L2
decremented.
Suppose, now, that we replace the INC EBX instruction above with another
jump instruction. This third jump instruction will then use the same
BTB entry as JMP L2 with the possible penalty of predicting a wrong target,
(unless it happens to also have L2 as target).
To summarize, consecutive jumps can lead to the following problems:
- failure to load a jump target when the pipeline is being flushed by a
preceding mispredicted jump.
- a BTB entry being mis-applied to non-jumping instructions and predicting
them to jump.
- a second consequence of the above is that a mis-applied BTB entry will
get its state decremented, possibly leading to a later misprediction of
the jump it belongs to. Even unconditional jumps can be predicted to fall
through for this reason.
- two jump instructions may share the same BTB entry, leading to the
prediction of a wrong target.
All this mess may give you a lot of penalties, so you should definitely
avoid having an instruction pair containing a jump immediately after another
poorly predictable jump.
It is time for an illustrative example:
CALL P
TEST EAX,EAX
JZ L2
L1: MOV [EDI],EBX
ADD EDI,4
DEC EAX
JNZ L1
L2: CALL P
This looks like a quite nice and normal piece of code: A function call,
a loop which is bypassed when the count is zero, and another function call.
How many problems can you spot in this program?
First, we may note that the function P is called alternatingly from two
different locations. This means that the target for the return from P will
be changing all the time. Consequently, the return from P will always be
mispredicted.
Assume, now, that EAX is zero. The jump to L2 will not have its target
loaded because the mispredicted return caused a pipeline flush. Next, the
second CALL P will also fail to have its target loaded because JZ L2
caused a pipeline flush. Here we have the situation where a chain of
consecutive jumps makes the pipeline flush repeatedly because the first
jump was mispredicted. The BTB entry for JZ L2 is stored at the address
of P's return instruction. This BTB entry will now be mis-applied to
whatever comes after the second CALL P, but that doesn't give a penalty
because the pipeline is flushed by the mispredicted second return.
Now, let's see what happens if EAX has a nonzero value the next time:
JZ L2 is always predicted to fall through because of the flush. The second
CALL P has a BTB entry at the address of TEST EAX,EAX. This entry will
be mis-applied to the MOV/ADD pair, predicting it to jump to P. This
causes a flush which prevents JNZ L1 from loading its target. If we have
been here before, then the second CALL P will have another BTB entry at
the address of DEC EAX. On the second and third iteration of the loop, this
entry will also be mis-applied to the MOV/ADD pair, until it has had its
state decremented to 1 or 0. This will not cause a penalty on the second
iteration because the flush from JNZ L1 prevents it from loading its false
target, but on the third iteration it will. The subsequent iterations of the
loop have no penalties, but when it exits, JNZ L1 is mispredicted. The
flush would now prevent CALL P from loading its target, were it not for
the fact that the BTB entry for CALL P has already been destroyed by
being mis-applied several times.
We can improve this code by putting in some NOP's to separate all consecutive
jumps:
CALL P
TEST EAX,EAX
NOP
JZ L2
L1: MOV [EDI],EBX
ADD EDI,4
DEC EAX
JNZ L1
L2: NOP
NOP
CALL P
The extra NOP's cost 2 clock cycles, but they save much more. Furthermore,
JZ L2 is now moved to the U-pipe which reduces its penalty from 4 to 3
when mispredicted. The only problem that remains is that the returns from
P are always mispredicted. This problem can only be solved by replacing the
call to P by an inline macro (if you have enough code cache).
The lesson to learn from this example is that you should always look
carefully for consecutive jumps and see if you can save time by inserting
some NOP's. You should be particularly aware of those situations where
misprediction is unavoidable, such as loop exits and returns from procedures
which are called from varying locations. If you have something useful to
put in, in stead of the NOP's, then you should of course do so.
Multiway branches (case statements) may be implemented either as a tree of
branch instructions or as a list of jump addresses. If you choose to use
a tree of branch instructions, then you have to include some NOP's or other
instructions to separate the consecutive branches. A list of jump addresses
may therefore be a better solution on the Pentium. The list of jump
addresses should be placed in the data segment. Never put data in the code
segment! On the PentiumPro you may choose the branch tree solution if it
gives a better prediction.
Multiple BTB entries
--------------------
While two jump instructions can share the same BTB entry, one jump
instruction can also have multiple BTB entries if it is jumped to from
different locations, or if its pairing is changing. Example:
JZ L1
MOV EAX,1
L1: MOV EBX,2
MOV ECX,3
JC L2
If JZ L1 falls through, then MOV EAX,1 pairs with MOV EBX,2, and
MOV ECX,3 pairs with JC L2. The BTB entry for JC L2 will be attached
to the address of MOV EAX,1. When JZ L1 is taken, then MOV EBX,2 will
pair with MOV ECX,3, JC L2 will be unpaired, and its BTB entry will be
at MOV EBX,2. So you have two BTB entries for JC L2. The first entry
is used when the zero flag is clear, the other when set. This may actually
be an advantage, because it will improve the predictability of the second
branch instruction if there is a correllation between the two. Assume, for
example, that this code sequence is preceded by a CMP instruction. Then
we can be certain that the zero flag and the carry flag will never both be
set at the same time. We will get no second BTB entry for JC L2 at
MOV EBX,2 because it always falls through if the zero flag is set, and
we will never get a misprediction for JC L2 when JZ L1 is taken.
The situations where you can take advantage of this trick are rare,
however, because you might as well let L1 go to another branch which
doesn't test the carry flag at all.
Tight loops
-----------
In a small loop you will often access the same BTB entry repeatedly with
small intervals. This never causes a stall. Rather than waiting for a BTB
entry to be updated, the Pentium somehow bypasses the pipeline and gets the
resulting state from the last jump before it has been written to the BTB.
This mechanism is almost transparent to the user, but it does in some cases
have funny effects: You can see a branch prediction going from state 0 to
state 1, rather than to state 3, if the zero has not yet been written to the
BTB. This happens if the loop has no more than four instruction pairs. In
loops with only two instruction pairs you may sometimes have state 0 for two
consecutive iterations without going out of the BTB. In such small loops it
also happens in rare cases that the prediction uses the state resulting from
two iterations ago, rather than from the last iteration. These funny effects
will usually not have any negative effects on performance.
Avoiding branches
-----------------
Sometimes it is possible to obtain the same effect as a branch by ingenious
manipulation of bits and flags. For example you may calculate the absoulte
value of a signed number without branching:
MOV EDX,EAX
SAR EDX,31
XOR EAX,EDX
SUB EAX,EDX
The carry flag is particularly useful for this kind of tricks.
Setting carry if a value is zero: CMP [value],1
Setting carry if a value is not zero: XOR EAX,EAX / CMP EAX,[value]
Incrementing a counter if carry: ADC EAX,0
Setting a bit for each time the carry is set: RCL EAX,1
Generating a bit mask if carry is set: SBB EAX,EAX
This example finds the minimum of two unsigned numbers: if (b < a) a = b;
SUB EBX,EAX
SBB ECX,ECX
AND ECX,EBX
ADD EAX,ECX
This example chooses between two numbers: if (a != 0) a = b; else a = c;
CMP EAX,1
SBB EAX,EAX
XOR ECX,EBX
AND EAX,ECX
XOR EAX,EBX
Whether or not such tricks are worth the extra code depend on how
predictable a conditional jump would be, whether the extra pairing
opportunities of the branch-free code can be utilized, and whether there
are other jumps following immediately after which could suffer the penalties
of consecutive jumps.
Later processors
----------------
The PentiumPro and MMX processors have a more advanced branch prediction
mechanism which enable them to correctly predict a simple repetitive
pattern. They do not have the asymmetry flaw that the Pentium without MMX
has. A conditional jump, which has not been seen before, is predicted
taken if it goes backwards (e.g. a loop), and not taken if it goes
forward, on these new processors.
MMX processors have a return stack buffer (RSB) which remembers where
calls come from, so you will no longer get mispredictions when calling
a subroutine from varying locations. To make this mechanism work, you
should always match calls with returns. This means that you should not
use the return instruction as an indirect jump or use an indirect jump
as return.
Mispredictions are very expensive on the PentiumPro (10-16 clocks!). You
should therefore avoid poorly predictable branches, and replace them with
conditional moves whenever possible.
The new processors also have problems with consecutive jumps, but for
different reasons. So avoiding consecutive jumps may be desirable on all
processors.
13. PREFIXES
============
An instruction with one or more prefixes may not be able to execute in the
V-pipe (se paragraph 8.7), and it may take more than one clock cycle to
decode. On the Pentium without MMX, the decoding delay is one clock cycle
for each prefix except for the 0Fh prefix of conditional near jumps.
The Pentium with MMX has no decoding delay for 0Fh prefix. Segment and repeat
prefixes take one clock extra to decode. Address and operand size prefixes
take two clocks extra to decode (one for reading the prefix, and one for
calculating the length of the instruction).
Address size prefixes can be avoided by using 32 bit mode.
Segment prefixes can be avoided in 32 bit mode by using a flat memory model.
Operand size prefixes can be avoided in 32 bit mode by using only 8 bit and
32 bit integers.
Where prefixes are unavoidable, the decoding delay may be masked if a
preceding instruction takes more than one clock cycle to execute. The rule
for the Pentium without MMX is that any instruction which takes N clock
cycles to execute (not to decode) can 'overshadow' the decoding delay of N-1
prefixes in the next two (sometimes three) instructions or instruction
pairs. In other words, each extra clock cycle that an instruction takes to
execute can be used to decode one prefix in a later instruction. This
shadowing effect even extends across a predicted branch. Any instruction
which takes more than one clock cycle to execute, and any instruction which
is delayed because of an AGI stall, cache miss, misalignment, or any other
reason except decoding delay and branch misprediction, has a shadowing
effect.
The Pentium with MMX has a similar shadowing effect, but the mechanism is
different. Decoded instructions are stored in a transparent
first-in-first-out (FIFO) buffer, which can hold up to four instructions.
As long as there are instructions in the FIFO buffer you get no delay.
When the buffer is empty then instructions are executed as soon as they
are decoded. The buffer is filled when instructions are decoded faster
than they are executed, i.e. when you have unpaired or multi-cycle
instructions. The FIFO buffer is emptied when instructions execute faster
than they are decoded, i.e. when you have decoding delays due to prefixes.
The FIFO buffer is empty after a mispredicted branch. The FIFO buffer can
receive two instructions per clock cycle provided that the second
instruction is without prefixes and none of the instructions are longer
than 7 bytes. The two execution pipelines (U and V) can each receive one
instruction per clock cycle from the FIFO buffer.
Examples:
CLD / REP MOVSD
The CLD instruction takes two clock cycles and can therefore overshadow
the decoding delay of the REP prefix. The code would take one clock cycle
more if the CLD instruction was placed far from the REP MOVSD.
CMP DWORD PTR [EBX],0 / MOV EAX,0 / SETNZ AL
The CMP instruction takes two clock cycles here because it is a read/modify
instruction. The 0FH prefix of the SETNZ instruction is decoded during the
second clock cycle of the CMP instruction, so that the decoding delay is
hidden.
14. REDUCING CODE SIZE
======================
As explained in paragraph 6, the code cache is 8 or 16 kb. If you have
problems keeping the critical parts of your code within the code cache,
then you may consider reducing the size of your code.
32 bit code is usually bigger than 16 bit code because addresses and data
constants take 4 bytes in 32 bit code and only 2 bytes in 16 bit code.
However, 16 bit code has other penalties such as prefixes and problems
with accessing adjacent words simultaneously (see chapter 10). Some other
methods for reducing the size or your code are discussed below.
Both jump addresses, data addresses, and data constants take less space if
they can be expressed as a sign-extended byte, i.e. if they are within the
interval from -128 to +127.
For jump addresses this means that short jumps take two bytes of code,
whereas jumps beyond 127 bytes take 5 bytes if unconditional and 6 bytes
if conditional.
Likewise, data addresses take less space if they can be expressed as a
pointer and a displacement between -128 and +127.
Example:
MOV EBX,DS:[100000] / ADD EBX,DS:[100004] ; 12 bytes
Reduce to:
MOV EAX,100000 / MOV EBX,[EAX] / ADD EBX,[EAX+4] ; 10 bytes
The advantage of using a pointer obviously increases if you use it many
times. Storing data on the stack and using EBP or ESP as pointer will thus
make your code smaller than if you use static memory locations and absolute
addresses, provided of course that your data are within 127 bytes of the
pointer. Using PUSH and POP to write and read temporary data is even more
advantageous.
Data constants may also take less space if they are between -128 and +127.
Most instructions with immediate operands have a short form where the
operand is a sign-extended single byte. Examples:
PUSH 200 ; 5 bytes
PUSH 100 ; 2 bytes
ADD EBX,128 ; 6 bytes
SUB EBX,-128 ; 3 bytes
The most important instruction with an immediate operand which doesn't
have such a short form is MOV. Examples:
MOV EAX, 1 ; 5 bytes
XOR EAX,EAX / INC EAX ; 3 bytes
PUSH 1 / POP EAX ; 3 bytes
If the same constant is used more than once then you may of course load it
into a register. Example:
MOV DWORD PTR [EBX],0 / MOV DWORD PTR [EBX+4],0 ; 13 bytes
XOR EAX,EAX / MOV [EBX],EAX / MOV [EBX+4],EAX ; 7 bytes
You may also consider that different instructions have different lengths.
The following instructions take only one byte and are therefore very
attractive: PUSH reg, POP reg, INC reg32, DEC reg32.
INC and DEC with 8 bit registers take 2 bytes, so INC EAX is shorter
than INC AL.
XCHG EAX,reg is also a single-byte instruction and thus takes less space
than MOV EAX,reg, but it is slower and not pairable.
Some instructions take one byte less when they use the accumulator than
when they use any other register. Examples:
MOV EAX,DS:[100000] is smaller than MOV EBX,DS:[100000]
ADD EAX,1000 is smaller than ADD EBX,1000
Instructions with pointers take one byte less when they have only a base
pointer (not ESP) and a displacement than when they have a scaled index
register, or both base pointer and index register, or ESP as base pointer.
Examples:
MOV EAX,[array][EBX] is smaller than MOV EAX,[array][EBX*4]
MOV EAX,[EBP+12] is smaller than MOV EAX,[ESP+12]
Instructions with EBP as base pointer and no displacement and no index
take one byte more than with other registers:
MOV EAX,[EBX] is smaller than MOV EAX,[EBP], but
MOV EAX,[EBX+4] is same size as MOV EAX,[EBP+4].
15. SCHEDULING FLOATING POINT CODE
==================================
Floating point instructions cannot pair the way integer instructions can,
except for one special case, defined by the following rules:
- the first instruction (executing in the U-pipe) must be FLD, FADD, FSUB,
FMUL, FDIV, FCOM, FCHS, or FABS
- the second instruction (in V-pipe) must be FXCH
- the instruction following the FXCH must be a floating point instruction,
otherwise the FXCH will take an extra clock cycle.
This special pairing is important, as will be explained shortly.
While floating point instructions in general cannot be paired, many can be
pipelined, i.e. one instruction can begin before the previous instruction has
finished. Example:
FADD ST(1),ST(0) ; clock cycle 1-3
FADD ST(2),ST(0) ; clock cycle 2-4
FADD ST(3),ST(0) ; clock cycle 3-5
FADD ST(4),ST(0) ; clock cycle 4-6
Obviously, two instructions cannot overlap if the second instruction needs
the result of the first. Since almost all floating point instructions
involve the top of stack register, ST(0), there are seemingly not very many
possibilities for making an instruction independent of the result of
previous instructions. The solution to this problem is register renaming.
The FXCH instruction does not in reality swap the contents of two registers,
it only swaps their names. Instructions which push or pop the register
stack also work by renaming. Register renaming has been highly optimized on
the Pentium so that a register may be renamed while in use. Register
renaming never causes stalls - it is even possible to rename a register more
than once in the same clock cycle, as for example when you pair FLD or
FCOMPP with FXCH.
By the proper use of FXCH instructions you may obtain a lot of overlapping in
your floating point code. Example:
FLD [a1] ; clock cycle 1
FADD [a2] ; clock cycle 2-4
FLD [b1] ; clock cycle 3
FADD [b2] ; clock cycle 4-6
FLD [c1] ; clock cycle 5
FADD [c2] ; clock cycle 6-8
FXCH ST(2) ; clock cycle 6
FADD [a3] ; clock cycle 7-9
FXCH ST(1) ; clock cycle 7
FADD [b3] ; clock cycle 8-10
FXCH ST(2) ; clock cycle 8
FADD [c3] ; clock cycle 9-11
FXCH ST(1) ; clock cycle 9
FADD [a4] ; clock cycle 10-12
FXCH ST(2) ; clock cycle 10
FADD [b4] ; clock cycle 11-13
FXCH ST(1) ; clock cycle 11
FADD [c4] ; clock cycle 12-14
FXCH ST(2) ; clock cycle 12
In the above example we are interleaving three independent threads. Each
FADD takes 3 clock cycles, and we can start a new FADD in each clock cycle.
When we have started an FADD in the 'a' thread we have time to start two new
FADD instructions in the 'b' and 'c' threads before returning to the 'a'
thread, so every third FADD belongs to the same thread. We are using FXCH
instructions every time to get the register that belongs to the desired
thread into ST(0). As you can see in the example above, this generates a
regular pattern, but note well that the FXCH instructions repeat with a
period of two while the threads have a period of three. This can be quite
confusing, so you have to 'play computer' in order to know which registers
are where.
All versions of the instructions FADD, FSUB, FMUL, and FILD take 3 clock
cycles and are able to overlap, so that these instructions may be scheduled
using the method described above. Using a memory operand does not take more
time than a register operand if the memory operand is in the level 1 cache
and properly aligned.
By now you must be used to the rules having exceptions, and the overlapping
rule is no exception: You cannot start an FMUL instruction one clock cycle
after another FMUL instruction, because the FMUL circuitry is not perfectly
pipelined. It is recommended that you put another instruction in between two
FMUL's. Example:
FLD [a1] ; clock cycle 1
FLD [b1] ; clock cycle 2
FLD [c1] ; clock cycle 3
FXCH ST(2) ; clock cycle 3
FMUL [a2] ; clock cycle 4-6
FXCH ; clock cycle 4
FMUL [b2] ; clock cycle 5-7 (stall)
FXCH ST(2) ; clock cycle 5
FMUL [c2] ; clock cycle 7-9 (stall)
FXCH ; clock cycle 7
FSTP [a3] ; clock cycle 8-9
FXCH ; clock cycle 10 (unpaired)
FSTP [b3] ; clock cycle 11-12
FSTP [c3] ; clock cycle 13-14
Here you have a stall before FMUL [b2] and before FMUL [c2] because
another FMUL started in the preceding clock cycle. You can improve this code
by putting FLD instructions in between the FMUL's:
FLD [a1] ; clock cycle 1
FMUL [a2] ; clock cycle 2-4
FLD [b1] ; clock cycle 3
FMUL [b2] ; clock cycle 4-6
FLD [c1] ; clock cycle 5
FMUL [c2] ; clock cycle 6-8
FXCH ST(2) ; clock cycle 6
FSTP [a3] ; clock cycle 7-8
FSTP [b3] ; clock cycle 9-10
FSTP [c3] ; clock cycle 11-12
In other cases you may put FADD, FSUB, or anything else in between FMUL's to
avoid the stalls.
Overlapping floating point instructions requires of course that you have
some independent threads that you can interleave. If you have only one big
formula to execute, then you may compute parts of the formula in parallel to
achieve overlapping. If, for example, you want to add six numbers, then you
may split the operations into two threads with three numbers in each, and
add the two threads in the end:
FLD [a] ; clock cycle 1
FADD [b] ; clock cycle 2-4
FLD [c] ; clock cycle 3
FADD [d] ; clock cycle 4-6
FXCH ; clock cycle 4
FADD [e] ; clock cycle 5-7
FXCH ; clock cycle 5
FADD [f] ; clock cycle 7-9 (stall)
FADD ; clock cycle 10-12 (stall)
Here we have a one clock stall before FADD [f] because it is waiting for
the result of FADD [d] and a two clock stall before the last FADD because
it is waiting for the result of FADD [f]. The latter stall can be hidden by
filling in some integer instructions, but the first stall can not because an
integer instruction at this place would make the FXCH unpairable.
The first stall can be avoided by having three threads rather than two, but
that would cost an extra FLD so we do not save anything by having three
threads rather than two unless there are at least eight numbers to add.
Not all floating point instructions can overlap. And some floating point
instructions can overlap more subsequent integer instructions than
subsequent floating point instructions. The FDIV instruction, for example,
takes 39 clock cycles. All but the first clock cycle can overlap with
integer instructions, but only the last two clock cycles can overlap with
floating point instructions. Example:
FDIV ; clock cycle 1-39
FXCH ; clock cycle 1-2
CMC ; clock cycle 3-4
RCR EAX,1 ; clock cycle 5
INC EBX ; clock cycle 5
FADD [x] ; clock cycle 38-40
FXCH ; clock cycle 38
FMUL [y] ; clock cycle 40-42
The first FXCH pairs with the FDIV, but takes an extra clock cycle because
it is not followed by a floating point instruction. The CMC starts before
the FDIV is finished, but has to wait for the FXCH to finish. The RCR and
INC instructions are pairing. The FADD has to wait till clock 38 because new
floating point instructions can only execute during the last two clock
cycles of the FDIV. The second FXCH pairs with the FADD. The FMUL has to
wait for the FDIV to finish because it uses the result of the division.
If you have nothing else to put in after a floating point instruction with a
large integer overlap, such as FDIV or FSQRT, then you may put in a dummy
read from an address which you expect to need later in the program to make
sure it is in the level one cache. Example:
FDIV QWORD PTR [EBX]
CMP [ESI],EAX
FMUL QWORD PTR [ESI]
Here we use the integer overlap to pre-load the value at [ESI] into the
cache while the FDIV is being computed (we don't care what the result of the
CMP is).
Paragraph 21 gives a complete listing of floating point instructions, and
what they can pair or overlap with.
There is no penalty for using a memory operand on floating point
instuctions because the arithmetic unit is one step later in the pipeline
than the read unit. The tradeoff of this comes when you store floating
point data to memory. The FST or FSTP instruction with a memory operand
takes two clock cycles in the execution stage, but it needs the data one
clock earlier so you will get a one clock stall if the value to store is
not ready one clock cycle in advance. This is analogous to an AGI stall.
Example:
FLD [a1] ; clock cycle 1
FADD [a2] ; clock cycle 2-4
FLD [b1] ; clock cycle 3
FADD [b2] ; clock cycle 4-6
FXCH ; clock cycle 4
FSTP [a3] ; clock cycle 6-7
FSTP [b3] ; clock cycle 8-9
The FSTP [a3] stalls for one clock cycle because the result of FADD [a2]
is not ready in the preceding clock cycle. In many cases you cannot hide
this type of stall without scheduling your floating point code into four
threads or putting some integer instructions in between.
The two clock cycles in the execution stage of the FST(P) instruction
cannot pair or overlap with any subsequent instructions.
Instructions with integer operands such as FIADD, FISUB, FIMUL, FIDIV, FICOM
may be split up into simpler operations in order to improve overlapping.
Example:
FILD [a] ; clock cycle 1-3
FIMUL [b] ; clock cycle 4-9
Split up into:
FILD [a] ; clock cycle 1-3
FILD [b] ; clock cycle 2-4
FMUL ; clock cycle 5-7
In this example, you save two clocks by overlapping the two FILD instructions.
16. Loop Optimation
===================
When analyzing a program you often find that most of the time consumption
lies in the innermost loop. The way to improve the speed is to carefully
optimize the most time-consuming loop using ASM language. The rest of the
program may be left in high-level language.
A loop generally contains a counter controlling how many times to iterate,
and often array access reading or writing one array element for each
iteration. I have chosen as example a procedure which reads integers from
an array, changes the sign of each integer, and stores the results in
another array.
A C language code for this procedure would be:
void ChangeSign (int * A, int * B, int N) {
int i;
for (i=0; i<N; i++) B[i] = -A[i];}
Translating to assembler, we might write the procedure like this:
Example 1:
_ChangeSign PROCEDURE NEAR
PUSH ESI
PUSH EDI
A EQU DWORD PTR [ESP+12]
B EQU DWORD PTR [ESP+16]
N EQU DWORD PTR [ESP+20]
MOV ECX, [N]
JECXZ L2
MOV ESI, [A]
MOV EDI, [B]
CLD
L1: LODSD
NEG EAX
STOSD
LOOP L1
L2: POP EDI
POP ESI
RET ; (no extra pop if C calling convention)
_ChangeSign ENDP
This looks like a nice solution, but it is not optimal because it uses
slow non-pairable instructions. It takes 11 clock cycles per iteration
if all data are in the level one cache.
Using pairable instructions only
--------------------------------
Example 2:
MOV ECX, [N]
MOV ESI, [A]
TEST ECX, ECX
JZ SHORT L2
MOV EDI, [B]
L1: MOV EAX, [ESI] ; u
XOR EBX, EBX ; v (pairs)
ADD ESI, 4 ; u
SUB EBX, EAX ; v (pairs)
MOV [EDI], EBX ; u
ADD EDI, 4 ; v (pairs)
DEC ECX ; u
JNZ L1 ; v (pairs)
L2:
Here we have used pairable instructions only, and scheduled the instruc-
tions so that everything pairs. It now takes only 4 clock cycles per
iteration. We could have obtained the same speed without splitting the
NEG instruction, but the other unpairable instructions should be split
up.
Using the same register for counter and index
---------------------------------------------
Example 3:
MOV ESI, [A]
MOV EDI, [B]
MOV ECX, [N]
XOR EDX, EDX
TEST ECX, ECX
JZ SHORT L2
L1: MOV EAX, [ESI+4*EDX] ; u
NEG EAX ; u
MOV [EDI+4*EDX], EAX ; u
INC EDX ; v (pairs)
CMP EDX, ECX ; u
JB L1 ; v (pairs)
L2:
Using the same register for counter and index gives us fewer instructions
in the body of the loop, but it still takes 4 clocks because we have two
unpaired instructions.
Letting the counter end at zero
-------------------------------
We want to get rid of the CMP instruction in example 3 by letting the
counter end at zero and use the zero flag for detecting when we are
finished as we did in example 2. One way to do this would be to execute
the loop backwards taking the last array elements first. However, data
caches are optimized for accessing data forwards, not backwards, so if
cache misses are likely, then you should rather start the counter at -N
and count through negative values up to zero. The base registers should
then point to the end of the arrays rather than the beginning:
Example 4:
MOV ESI, [A]
MOV EAX, [N]
MOV EDI, [B]
XOR ECX, ECX
LEA ESI, [ESI+4*EAX] ; point to end of array A
SUB ECX, EAX ; -N
LEA EDI, [EDI+4*EAX] ; point to end of array B
JZ SHORT L2
L1: MOV EAX, [ESI+4*ECX] ; u
NEG EAX ; u
MOV [EDI+4*ECX], EAX ; u
INC ECX ; v (pairs)
JNZ L1 ; u
L2:
We are now down at five instructions in the loop body but it still takes
4 clocks because of poor pairing. (If the addresses and sizes of the
arrays are constants we may save two registers by substituting A+SIZE A
for ESI and B+SIZE B for EDI). Now let's see how we can improve pairing.
Pairing calculations with loop overhead
---------------------------------------
We may want to improve pairing by intermingling calculations with the loop
control instructions. If we want to put something in between INC ECX and
JNZ L1, it has to be something that doesn't affect the zero flag. The
MOV [EDI+4*ECX],EBX instruction after INC ECX would generate an AGI
delay, so we have to be more ingenious:
Example 5:
MOV EAX, [N]
XOR ECX, ECX
SHL EAX, 2 ; 4 * N
JZ SHORT L3
MOV ESI, [A]
MOV EDI, [B]
SUB ECX, EAX ; - 4 * N
ADD ESI, EAX ; point to end of array A
ADD EDI, EAX ; point to end of array B
JMP SHORT L2
L1: MOV [EDI+ECX-4], EAX ; u
L2: MOV EAX, [ESI+ECX] ; v (pairs)
XOR EAX, -1 ; u
ADD ECX, 4 ; v (pairs)
INC EAX ; u
JNC L1 ; v (pairs)
MOV [EDI+ECX-4], EAX
L3:
I have used a different way to calculate the negative of EAX here:
inverting all bits and adding one. The reason why I am using this method
is that I can use a dirty trick with the INC instruction: INC doesn't
change the carry flag, whereas ADD does. I am using ADD rather than INC
to increment my loop counter and testing the carry flag rather than the
zero flag. It is then possible to put the INC EAX in between without
affecting the carry flag. You may think that we could have used
LEA EAX,[EAX+1] here instead of INC EAX, at least that doesn't change
any flags, but the LEA instruction would have an AGI stall so that's not
the best solution.
I have obtained perfect pairing here and the loop now takes only 3 clock
cycles. Whether you want to increment the loop counter by 1 (as in example
4) or by 4 (as in example 5) is a matter of taste, it makes no difference
in loop timing.
Overlapping the end of one operation with the beginning of the next
-------------------------------------------------------------------
The method used in example 5 is not very generally applicable so we may
look for other methods of improving pairing opportunities. One way is to
reorganize the loop so that the end of one operation overlaps with the
beginning of the next. I will call this convoluting the loop. A convoluted
loop has an unfinished operation at the end of each loop iteration which
will be finished in the next run. Actually, example 5 did pair the last
MOV of one iteration with the first MOV of the next, but we want to
explore this method further:
Example 6:
MOV ESI, [A]
MOV EAX, [N]
MOV EDI, [B]
XOR ECX, ECX
LEA ESI, [ESI+4*EAX] ; point to end of array A
SUB ECX, EAX ; -N
LEA EDI, [EDI+4*EAX] ; point to end of array B
JZ SHORT L3
XOR EBX, EBX
MOV EAX, [ESI+4*ECX]
INC ECX
JZ SHORT L2
L1: SUB EBX, EAX ; u
MOV EAX, [ESI+4*ECX] ; v (pairs)
MOV [EDI+4*ECX-4], EBX ; u
INC ECX ; v (pairs)
MOV EBX, 0 ; u
JNZ L1 ; v (pairs)
L2: SUB EBX, EAX
MOV [EDI+4*ECX-4], EBX
L3:
Here we begin reading the second value before we have stored the first,
and this of course improves pairing opportunities. The MOV EBX,0
instruction has been put in between INC ECX and JNZ L1 not to improve
pairing but to avoid AGI stall.
Rolling out a loop
------------------
The most generally applicable way to improve pairing opportunities is to
do two operations for each run and do half as many runs. This is called
rolling out a loop:
Example 7:
MOV ESI, [A]
MOV EAX, [N]
MOV EDI, [B]
XOR ECX, ECX
LEA ESI, [ESI+4*EAX] ; point to end of array A
SUB ECX, EAX ; -N
LEA EDI, [EDI+4*EAX] ; point to end of array B
JZ SHORT L2
TEST AL,1 ; test if N is odd
JZ SHORT L1
MOV EAX, [ESI+4*ECX] ; N is odd. do the odd one
NEG EAX
MOV [EDI+4*ECX], EAX
INC ECX ; make counter even
JZ SHORT L2 ; N = 1
L1: MOV EAX, [ESI+4*ECX] ; u
MOV EBX, [ESI+4*ECX+4] ; v (pairs)
NEG EAX ; u
NEG EBX ; u
MOV [EDI+4*ECX], EAX ; u
MOV [EDI+4*ECX+4], EBX ; v (pairs)
ADD ECX, 2 ; u
JNZ L1 ; v (pairs)
L2:
Now we are doing two operations in parallel which gives the best pairing
opportunities. We have to test if N is odd and if so do one operation
outside the loop because the loop can only do an even number of operations.
The loop has an AGI stall at the first MOV instruction because ECX has
been incremented in the preceding clock cycle. The loop therefore takes
6 clock cycles for two operations.
Reorganizing a loop to remove AGI stall
---------------------------------------
Example 8:
MOV ESI, [A]
MOV EAX, [N]
MOV EDI, [B]
XOR ECX, ECX
LEA ESI, [ESI+4*EAX] ; point to end of array A
SUB ECX, EAX ; -N
LEA EDI, [EDI+4*EAX] ; point to end of array B
JZ SHORT L3
TEST AL,1 ; test if N is odd
JZ SHORT L2
MOV EAX, [ESI+4*ECX] ; N is odd. do the odd one
NEG EAX ; no pairing opportunity
MOV [EDI+4*ECX-4], EAX
INC ECX ; make counter even
JNZ SHORT L2
NOP ; add NOP's if JNZ L2 not predictable
NOP
JMP SHORT L3 ; N = 1
L1: NEG EAX ; u
NEG EBX ; u
MOV [EDI+4*ECX-8], EAX ; u
MOV [EDI+4*ECX-4], EBX ; v (pairs)
L2: MOV EAX, [ESI+4*ECX] ; u
MOV EBX, [ESI+4*ECX+4] ; v (pairs)
ADD ECX, 2 ; u
JNZ L1 ; v (pairs)
NEG EAX
NEG EBX
MOV [EDI+4*ECX-8], EAX
MOV [EDI+4*ECX-4], EBX
L3:
The trick is to find a pair of instructions that do not use the loop
counter as index and reorganize the loop so that the counter is
incremented in the preceding clock cycle. We are now down at 5 clock
cycles for two operations which is close to the best possible.
If data caching is critical, then you may improve the speed further by
interleaving the A and B arrays into one structured array so that each
B[i] comes immediately after the corresponding A[i]. If the structured
array is aligned by at least 8 then B[i] will always be in the same cache
line as A[i], so you will never have a cache miss when writing B[i]. This
may of course have a tradeoff in other parts of the program so you have
to weigh the costs against the benefits.
Rolling out by more than 2
--------------------------
You may think of doing more than two operations per iteration in order to
reduce the loop overhead per operation. But since the loop overhead in
most cases can be reduced to only one clock cycle per iteration, then
rolling out the loop by 4 rather than by 2 would only save 1/4 clock cycle
per operation, which is hardly worth the effort. Only if the loop overhead
cannot be reduced to one clock cycle and if N is very big, should you
think of unrolling by 4.
The drawbacks of excessive loop unrolling are:
1. You need to calculate N MODULO R, where R is the unrolling factor,
and do N MODULO R operations before or after the main loop in order
to make the remaining number of operations divisible by R. This takes
a lot of extra code and poorly predictable branches. And the loop body
of course also becomes bigger.
2. A Piece of code usually takes much more time the first time it
executes, and the penalty of first time execution is bigger the more
code you have, especially if N is small.
3. Excessive code size makes the utilization of the code cache less
effective.
Handling multiple 8 or 16 bit operands simultaneously in 32 bit registers
-------------------------------------------------------------------------
If you need to manipulate arrays of 8 or 16 bit operands, then there is a
problem with unrolled loops because you may not be able to pair two memory
access operations. For example MOV AL,[ESI] / MOV BL,[ESI+1] will not pair
if the two operands are within the same dword of memory. But there may be a
much smarter method, namely to handle four bytes at a time in the same 32
bit register. The MMX processors have special instructions for exactly this
purpose, but if you want your code to run on non-MMX processors or if you
have floating point instructions in the same part of the program so that
you cannot use MMX instructions, then you may want to know how to handle
this in the general 32 bit registers.
The following example adds 2 to all elements of an array of bytes:
Example 9:
MOV ESI, [A] ; address of byte array
MOV ECX, [N] ; number of elements in byte array
TEST ECX, ECX ; test if N is 0
JZ SHORT L2
MOV EAX, [ESI] ; read first four bytes
L1: MOV EBX, EAX ; copy into EBX
AND EAX, 7F7F7F7FH ; get lower 7 bits of each byte in EAX
XOR EBX, EAX ; get the highest bit of each byte in EBX
ADD EAX, 02020202H ; add desired value to all four bytes
XOR EBX, EAX ; combine bits again
MOV EAX, [ESI+4] ; read next four bytes
MOV [ESI], EBX ; store result
ADD ESI, 4 ; increment pointer
SUB ECX, 4 ; decrement loop counter
JA L1 ; loop
L2:
This loop takes 5 clock cycles for every 4 bytes. The array should of course
be aligned by 4. If the number of elements in the array is not divisible by
four, then you may pad it in the end with a few extra bytes to make the
length divisible by four. This loop will always read past the end of the
array, so you should make sure the array is not placed at the end of a
segment to avoid a general protection error.
Note that I have masked out the highest bit of each byte to avoid a possible
carry from each byte into the next when adding. I am using XOR rather than
ADD when putting in the high bit again to avoid carry.
The ADD ESI,4 instruction could have been avoided by using the loop
counter as index as in example 4. However, this would give an odd number of
instructions in the loop body, so there would be one unpaired instruction
and the loop would still take 5 clocks. Making the branch instruction
unpaired would save one clock after the last operation when the branch is
mispredicted, but we would have to spend an extra clock cycle in the prolog
code to setup a pointer to the end of the array and calculate -N, so the two
methods will be exactly equally fast. The method presented here is the
simplest and shortest.
The next example finds the length of a zero-terminated string by searching
for the first byte of zero. It is faster than using REP SCASB:
Example 10:
STRLEN PROC NEAR
MOV EAX,[ESP+4] ; get pointer
MOV EDX,7
ADD EDX,EAX ; pointer+7 used in the end
PUSH EBX
MOV EBX,[EAX] ; read first 4 bytes
ADD EAX,4 ; increment pointer
L1: LEA ECX,[EBX-01010101H] ; subtract 1 from each byte
XOR EBX,-1 ; invert all bytes
AND ECX,EBX ; and these two
MOV EBX,[EAX] ; read next 4 bytes
ADD EAX,4 ; increment pointer
AND ECX,80808080H ; test all sign bits
JZ L1 ; no zero bytes, continue loop
TEST ECX,00008080H ; test first two bytes
JNZ SHORT L2
SHR ECX,16 ; not in the first 2 bytes
ADD EAX,2
L2: SHL CL,1 ; use carry flag to avoid a branch
POP EBX
SBB EAX,EDX ; compute length
RET ; (or RET 4 if pascal)
STRLEN ENDP
Again we have used the method of overlapping the end of one operation with
the beginning of the next to improve pairing. I have not unrolled the loop
because it is likely to repeat relatively few times. The string should of
course be aligned by 4. The code will always read past the end of the
string, so the string should not be placed at the end of a segment.
The loop body has an odd number of instructions so there is one unpaired.
Making the branch instruction unpaired rather than one of the other
instructions has the advantage that it saves 1 clock cycle when the branch
is mispredicted.
The TEST ECX,00008080H instruction is non-pairable. You could use the
pairable instruction OR CH,CL here instead, but then you would have to
put in a NOP or something to avoid the penalties of consequtive branches.
Another problem with OR CH,CL is that it would cause a partial register
stall on a 486 or PentiumPro processor. So I have chosen to keep the
unpairable TEST instruction.
Handling 4 bytes simultaneously can be quite difficult. The code uses a
formula which generates a nonzero value for a byte if, and only if, the
byte is zero. This makes it possible to test all four bytes in one
operation. This algorithm involves the subtraction of 1 from all bytes (in
the LEA instruction). I have not masked out the highest bit of each byte
before subtracting, as I did in the previous example, so the subtraction may
generate a borrow to the next byte, but only if it is zero, and this is
exactly the situation where we don't care what the next byte is, because we
are searching forwards for the first zero. If we were searching backwards
then we would have to re-read the dword after detecting a zero, and then
test all four bytes to find the last zero, or use BSWAP to reverse the order
of the bytes.
If you want to search for a byte value other than zero, then you may XOR all
four bytes with the value you are searching for, and then use the method
above to search for zero.
Handling multiple operands in the same register is easier on the MMX
processors because they have special instructions and special 64 bit
registers for this purpose. Using the MMX instructions has a high penalty if
you are using floating point instructions shortly afterwards, so there may
still be situations where you want to use 32 bit registers as in the
examples above.
Loops with floating point operations
------------------------------------
The methods of optimizing floating point loops are basically the same as
for integer loops, although the floating point instructions are
overlapping rather than pairing.
Consider the C language code:
int i, n; double * X; double * Y; double DA;
for (i=0; i<n; i++) Y[i] = Y[i] - DA * X[i];
This piece of code (called DAXPY) has been studied extensively because
it is the key to solving linear equations.
Example 11:
DSIZE = 8 ; data size
MOV EAX, [N] ; number of elements
MOV ESI, [X] ; pointer to X
MOV EDI, [Y] ; pointer to Y
XOR ECX, ECX
LEA ESI, [ESI+DSIZE*EAX] ; point to end of X
SUB ECX, EAX ; -N
LEA EDI, [EDI+DSIZE*EAX] ; point to end of Y
JZ SHORT L3 ; test for N = 0
FLD DSIZE PTR [DA]
FMUL DSIZE PTR [ESI+DSIZE*ECX] ; DA * X[0]
JMP SHORT L2 ; jump into loop
L1: FLD DSIZE PTR [DA]
FMUL DSIZE PTR [ESI+DSIZE*ECX] ; DA * X[i]
FXCH ; get old result
FSTP DSIZE PTR [EDI+DSIZE*ECX-DSIZE] ; store Y[i]
L2: FSUBR DSIZE PTR [EDI+DSIZE*ECX] ; subtract from Y[i]
INC ECX ; increment index
JNZ L1 ; loop
FSTP DSIZE PTR [EDI+DSIZE*ECX-DSIZE] ; store last result
L3:
Here we are using the same methods as in example 6: Using the loop counter
as index register and counting through negative values up to zero. The end
of one operation overlaps with the beginning of the next.
The interleaving of floating point operations work perfectly here: The 2
clock stall between FMUL and FSUBR is filled with the FSTP of the previous
result. The 3 clock stall between FSUBR and FSTP is filled with the loop
overhead and the first two instructions of the next operation. An AGI stall
has been avoided by reading the only parameter that doesn't depend on the
index in the first clock cycle after the index has been incremented.
This solution takes 6 clock cycles per operation, which is better than the
unrolled solution published by Intel!
Unrolling floating point loops
------------------------------
The DAXPY loop unrolled by 3 is quite complicated:
Example 12:
DSIZE = 8 ; data size
IF DSIZE EQ 4
SHIFTCOUNT = 2
ELSE
SHIFTCOUNT = 3
ENDIF
MOV EAX, [N] ; number of elements
MOV ECX, 3*DSIZE ; counter bias
SHL EAX, SHIFTCOUNT ; DSIZE*N
JZ L4 ; N = 0
MOV ESI, [X] ; pointer to X
SUB ECX, EAX ; (3-N)*DSIZE
MOV EDI, [Y] ; pointer to Y
SUB ESI, ECX ; end of pointer - bias
SUB EDI, ECX
TEST ECX, ECX
FLD DSIZE PTR [ESI+ECX] ; first X
JNS SHORT L2 ; less than 4 operations
L1: ; main loop
FMUL DSIZE PTR [DA]
FLD DSIZE PTR [ESI+ECX+DSIZE]
FMUL DSIZE PTR [DA]
FXCH
FSUBR DSIZE PTR [EDI+ECX]
FXCH
FLD DSIZE PTR [ESI+ECX+2*DSIZE]
FMUL DSIZE PTR [DA]
FXCH
FSUBR DSIZE PTR [EDI+ECX+DSIZE]
FXCH ST(2)
FSTP DSIZE PTR [EDI+ECX]
FSUBR DSIZE PTR [EDI+ECX+2*DSIZE]
FXCH
FSTP DSIZE PTR [EDI+ECX+DSIZE]
FLD DSIZE PTR [ESI+ECX+3*DSIZE]
FXCH
FSTP DSIZE PTR [EDI+ECX+2*DSIZE]
ADD ECX, 3*DSIZE
JS L1 ; loop
L2: FMUL DSIZE PTR [DA] ; finish leftover operation
FSUBR DSIZE PTR [EDI+ECX]
SUB ECX, 2*DSIZE ; change pointer bias
JZ SHORT L3 ; finished
FLD DSIZE PTR [DA] ; start next operation
FMUL DSIZE PTR [ESI+ECX+3*DSIZE]
FXCH
FSTP DSIZE PTR [EDI+ECX+2*DSIZE]
FSUBR DSIZE PTR [EDI+ECX+3*DSIZE]
ADD ECX, 1*DSIZE
JZ SHORT L3 ; finished
FLD DSIZE PTR [DA]
FMUL DSIZE PTR [ESI+ECX+3*DSIZE]
FXCH
FSTP DSIZE PTR [EDI+ECX+2*DSIZE]
FSUBR DSIZE PTR [EDI+ECX+3*DSIZE]
ADD ECX, 1*DSIZE
L3: FSTP DSIZE PTR [EDI+ECX+2*DSIZE]
L4:
The reason why I am showing you how to unroll a loop by 3 is not to
recommend it, but to warn you how difficult it is! Be prepared to spend a
considerable amount of time debugging and verifying your code when doing
something like this. There are several problems to take care of: In most
cases, you cannot remove all stalls from a floating point loop unrolled by
less than 4 unless you convolute it (i.e. there are unfinished operations
at the end of each run which are being finished in the next run). The last
FLD in the main loop above is the beginning of the first operation in the
next run. It would be tempting here to make a solution which reads past
the end of the array and then discards the extra value in the end, as in
example 9 and 10, but that is not recommended in floating point loops
because the reading of the extra value might generate a denormal operand
exception in case the memory position after the array doesn't contain a
valid floating point number. To avoid this, we have to do at least one
more operation after the main loop.
The number of operations to do outside an unrolled loop would normally be
N MODULO R, where N is the number of operations, and R is the unrolling
factor. But in the case of a convoluted loop, we have to do one more,
i.e. (N-1) MODULO R + 1, for the abovementioned reason.
Normally, we would prefer to do the extra operations before the main loop,
but here we have to do them afterwards for two reasons: One reason is to
take care of the leftover operand from the convolution. The other reason
is that calculating the number of extra operations requires a division if
R is not a power of 2, and a division is time consuming. Doing the extra
operations after the loop saves the division.
The next problem is to calculate how to bias the loop counter so that it
will change sign at the right time, and adjust the base pointers so as to
compensate for this bias. Finally, you have to make sure the leftover
operand from the convolution is handled correctly for all values of N.
The epilog code doing 1-3 operations could have been implemented as a
separate loop, but that would cost an extra branch misprediction, so the
solution above is faster.
Now that I have scared you by demonstrating how difficult it is to unroll
by 3, I will show you that it is much easier to unroll by 4:
Example 13:
DSIZE = 8 ; data size
MOV EAX, [N] ; number of elements
MOV ESI, [X] ; pointer to X
MOV EDI, [Y] ; pointer to Y
XOR ECX, ECX
LEA ESI, [ESI+DSIZE*EAX] ; point to end of X
SUB ECX, EAX ; -N
LEA EDI, [EDI+DSIZE*EAX] ; point to end of Y
TEST AL,1 ; test if N is odd
JZ SHORT L1
FLD DSIZE PTR [DA] ; do the odd operation
FMUL DSIZE PTR [ESI+DSIZE*ECX]
FSUBR DSIZE PTR [EDI+DSIZE*ECX]
INC ECX ; adjust counter
FSTP DSIZE PTR [EDI+DSIZE*ECX-DSIZE]
L1: TEST AL,2 ; test for possibly 2 more operations
JZ L2
FLD DSIZE PTR [DA] ; N MOD 4 = 2 or 3. Do two more
FMUL DSIZE PTR [ESI+DSIZE*ECX]
FLD DSIZE PTR [DA]
FMUL DSIZE PTR [ESI+DSIZE*ECX+DSIZE]
FXCH
FSUBR DSIZE PTR [EDI+DSIZE*ECX]
FXCH
FSUBR DSIZE PTR [EDI+DSIZE*ECX+DSIZE]
FXCH
FSTP DSIZE PTR [EDI+DSIZE*ECX]
FSTP DSIZE PTR [EDI+DSIZE*ECX+DSIZE]
ADD ECX, 2 ; counter is now divisible by 4
L2: TEST ECX, ECX
JZ L4 ; no more operations
L3: ; main loop:
FLD DSIZE PTR [DA]
FLD DSIZE PTR [ESI+DSIZE*ECX]
FMUL ST,ST(1)
FLD DSIZE PTR [ESI+DSIZE*ECX+DSIZE]
FMUL ST,ST(2)
FLD DSIZE PTR [ESI+DSIZE*ECX+2*DSIZE]
FMUL ST,ST(3)
FXCH ST(2)
FSUBR DSIZE PTR [EDI+DSIZE*ECX]
FXCH ST(3)
FMUL DSIZE PTR [ESI+DSIZE*ECX+3*DSIZE]
FXCH
FSUBR DSIZE PTR [EDI+DSIZE*ECX+DSIZE]
FXCH ST(2)
FSUBR DSIZE PTR [EDI+DSIZE*ECX+2*DSIZE]
FXCH
FSUBR DSIZE PTR [EDI+DSIZE*ECX+3*DSIZE]
FXCH ST(3)
FSTP DSIZE PTR [EDI+DSIZE*ECX]
FSTP DSIZE PTR [EDI+DSIZE*ECX+2*DSIZE]
FSTP DSIZE PTR [EDI+DSIZE*ECX+DSIZE]
FSTP DSIZE PTR [EDI+DSIZE*ECX+3*DSIZE]
ADD ECX, 4 ; increment index by 4
JNZ L3 ; loop
L4:
It is usually quite easy to find a stall-free solution when unrolling by
4, and there is no need for convolution. The number of extra operations
to do outside the main loop is N MODULO 4, which can be calculated easily
without division, simply by testing the two lowest bits in N. The extra
operations are done before the main loop rather than after, to make the
handling of the loop counter simpler.
The tradeoff of loop unrolling is that the extra operations outside the
loop are slower due to incomplete overlapping and possible branch
mispredictions, and the first time penalty is higher because of increased
code size.
As a general recommendation, I would say that if N is big or if
convoluting the loop without unrolling cannot remove enough stalls, then
you should unroll critical integer loops by 2 and floating point loops
by 4.
17. DISCUSSION OF SPECIAL INSTRUCTIONS
======================================
17.1 LEA
--------
The LEA instruction is useful for many purposes because it can do a shift,
two additions, and a move in just one pairable instruction taking one clock
cycle. Example:
LEA EAX,[EBX+8*ECX-1000]
is much faster than
MOV EAX,ECX / SHL EAX,3 / ADD EAX,EBX / SUB EAX,1000
The LEA instruction can also be used to do an add or shift without changing
the flags. The source and destination need not have the same word size, so
LEA EAX,[BX] is a useful replacement for MOVZX EAX,BX.
You must be aware, however, that the LEA instruction will suffer an AGI
stall if it uses a base or index register which has been changed in the
preceding clock cycle.
Since the LEA instruction is pairable in the V-pipe and shift instructions
are not, you may use LEA as a substitute for a SHL by 1, 2, or 3 if you want
the instruction to execute in the V-pipe.
The 32 bit processors have no documented addressing mode with a scaled index
register and nothing else, so an instruction like LEA EAX,[EAX*2] is
actually coded as LEA EAX,[EAX*2+00000000] with an immediate displacement
of 4 bytes. You may reduce the instruction size by instead writing LEA
EAX,[EAX+EAX] or even better ADD EAX,EAX. The latter code cannot have an
AGI delay. If you happen to have a register which is zero (like a loop
counter after a loop), then you may use it as a base register to reduce the
code size:
LEA EAX,[EBX*4] ; 7 bytes
LEA EAX,[ECX+EBX*4] ; 3 bytes
17.2 MOV [MEM],ACCUM
--------------------
The instructions MOV [mem],AL MOV [mem],AX MOV [mem],EAX
are treated by the pairing circuitry as if they were writing to the
accumulator. Thus the following instructions do not pair:
MOV [mydata], EAX
MOV EBX, EAX
This problem occurs only with the short form of the MOV instruction which
can not have a base or index register, and which can only have the
accumulator as source. You can avoid the problem by using another register,
by reordering your instructions, by using a pointer, or by hard-coding the
general form of the MOV instruction.
In 32 bit mode you can write the general form of MOV [mem],EAX:
DB 89H, 05H
DD OFFSET DS:mem
In 16 bit mode you can write the general form of MOV [mem],AX:
DB 89H, 06H
DW OFFSET DS:mem
To use AL instead of (E)AX, you replace 89H with 88H
This bug has not been fixed in the MMX version.
17.3 TEST
---------
The TEST instruction with an immediate operand is only pairable if the
destination is AL, AX, or EAX.
TEST register,register and TEST register,memory is always pairable.
Examples:
TEST ECX,ECX ; pairable
TEST [mem],EBX ; pairable
TEST EDX,256 ; not pairable
TEST DWORD PTR [EBX],8000H ; not pairable
To make it pairable, use any of the following methods:
MOV EAX,[EBX] / TEST EAX,8000H
MOV EDX,[EBX] / AND EDX,8000H
MOV AL,[EBX+1] / TEST AL,80H
MOV AL,[EBX+1] / TEST AL,AL ; (result in sign flag)
It is also possible to test a bit by shifting it into the carry flag:
MOV EAX,[EBX] / SHR EAX,16 ; (result in carry flag)
but this method has a penalty on the PentiumPro and PentiumII when the
shift count is more than one.
(The reason for this non-pairability is probably that the first byte of
the 2-byte instruction is the same as for some other non-pairable
instructions, and the Pentium cannot afford to check the second byte too
when determining pairability.)
17.4 XCHG
---------
The XCHG register,[memory] instruction is dangerous. By default this
instruction has an implicit LOCK prefix which prevents it from using the
cache. The instruction is therefore very time consuming, and should always
be avoided.
17.5 rotates through carry
--------------------------
RCR and RCL with a count different from one are slow and should be avoided.
17.6 string instructions
------------------------
String instructions without a repeat prefix are too slow, and should always
be replaced by simpler instructions. The same applies to LOOP and JECXZ.
String instructions with repeat may be optimal. Always use the dword version
if possible, and make sure that both source and destination are aligned by 4.
REP MOVSD is the fastest way to move blocks of data when the destination
is in the level 1 cache. See section 19 for an alternative. Note that
while the REP MOVS instruction writes a word to the destination, it reads
the next word from the source in the same clock cycle. According to rule
10.3 above, you have a cache bank conflict if bit 2-4 are the same in
these two addresses. In other words, you will get a penalty of one clock
extra per iteration if ESI+(wordsize)-EDI is divisible by 32. The easiest
way to avoid cache bank conflicts is to use the DWORD version and align
both source and destination by 8. Never use MOVSB or MOVSW in optimized
code, not even in 16 bit mode.
REP STOSD is optimal when the destination is in the cache.
REP LOADS, REP SCAS, and REP CMPS are not optimal, and may be replaced by
loops. See section 16 example 10 for an alternative to REP SCASB.
REP CMPS may suffer cache bank conflicts if bit 2-4 are the same in ESI
and EDI.
17.7 bit scan
-------------
BSF and BSR are the poorest optimized instructions on the Pentium, taking
approximately 11 + 2*n clock cycles, where n is the number of zeros skipped.
(on later processors it takes only 1 or 2)
The following code emulates BSR ECX,EAX:
TEST EAX,EAX
JZ SHORT BS1
MOV DWORD PTR [TEMP],EAX
MOV DWORD PTR [TEMP+4],0
FILD QWORD PTR [TEMP]
FSTP QWORD PTR [TEMP]
WAIT ; WAIT only needed for compatibility with old processors
MOV ECX, DWORD PTR [TEMP+4]
SHR ECX,20 ; isolate exponent
SUB ECX,3FFH ; adjust
TEST EAX,EAX ; clear zero flag
BS1:
The following code emulates BSF ECX,EAX:
TEST EAX,EAX
JZ SHORT BS2
XOR ECX,ECX
MOV DWORD PTR [TEMP+4],ECX
SUB ECX,EAX
AND EAX,ECX
MOV DWORD PTR [TEMP],EAX
FILD QWORD PTR [TEMP]
FSTP QWORD PTR [TEMP]
WAIT ; WAIT only needed for compatibility with earlier processors
MOV ECX, DWORD PTR [TEMP+4]
SHR ECX,20
SUB ECX,3FFH
TEST EAX,EAX ; clear zero flag
BS2:
17.8 bit test
-------------
BT, BTC, BTR, and BTS instructions should preferably be replaced by
instructions like TEST, AND, OR, XOR, or shifts.
17.9 integer multiplication
---------------------------
An integer multiplication takes approximately 9 clock cycles. It is
therefore advantageous to replace a multiplication by a constant with a
combination of other instructions such as SHL, ADD, SUB, and LEA.
Example:
IMUL EAX,10
can be replaced with
MOV EBX,EAX / ADD EAX,EAX / SHL EBX,3 / ADD EAX,EBX
or
LEA EAX,[EAX+4*EAX] / ADD EAX,EAX
Floating point multiplication is faster than integer multiplication on a
pentium with or without MMX, but the time used to convert integers to
float and convert the product back again is usually more than the time
saved by using floating point multiplication, except when the number of
conversions is low compared with the number of multiplications. MMX
multiplication is fast, but is only available with 16-bit operands.
17.10 division
--------------
Division is quite time consuming. The DIV instruction takes 17, 25, or 41
clock cycles for byte, word, and dword divisors respectively. The IDIV
instruction takes 5 clock cycles more. It is therefore preferable to use
the smallest operand size possible that won't generate an overflow, even
if it costs an operand size prefix, and use unsigned division if possible.
Division by a constant
----------------------
Unsigned division by a power of two can be done with SHR. Division of a
signed number by a power of two can be done with SAR, but the result with
SAR is rounded towards minus infinity, whereas the result with IDIV is
truncated towards zero.
Dividing by a constant can be done by multiplying with the reciprocal.
To calculate q = x / d, you first calculate the reciprocal f = 2^r / d,
where r defines the position of the binary decimal point (radix point).
Then multiply x with f and shift right r positions. The maximum value
of r is 32+b, where b is the number of binary digits in d minus 1.
(b is the highest integer for which 2^b <= d). Use r = 32+b to cover
the maximum range for the value of the dividend x.
This method needs some refinement in order to compensate for rounding
errors. The following algorithm will give you the correct result for
unsigned integer division with truncation, i.e. the same as the DIV
instruction does:
b = the number of significant bits in d - 1
r = 32 + b
f = 2^r / d
If f is an integer then d is a power of 2: goto case A.
If f is not an integer, then check if the fractional part of f is < 0.5
If the fractional part of f < 0.5: goto case B.
If the fractional part of f > 0.5: goto case C.
case A: (d = 2^b)
result = x SHR b
case B: (fractional part of f < 0.5)
round f down to nearest integer
result = ((x+1) * f) SHR r
case C: (fractional part of f > 0.5)
round f up to nearest integer
result = (x * f) SHR r
Example:
Assume that you want to divide by 5.
5 = 00000101b.
b = number of significant binary digits - 1 = 2
r = 32+2 = 34
f = 2^34 / 5 = 3435973836.8 = 0CCCCCCCC.CCC...(hexadecimal)
The fractional part is greater than a half: use case C.
Round f up to 0CCCCCCCDh.
This code divides EAX by 5 and returns the result in EDX:
MOV EDX,0CCCCCCCDh
MUL EDX
SHR EDX,2
After the multiplication, EDX contains the product shifted right 32
places. Since r = 34 you have to shift 2 more places to get the result.
To divide by 10 you just change the last line to SHR EDX,3.
In case B you would have:
INC EAX
MOV EDX,f
MUL EDX
SHR EDX,b
This code works for all values of x except 0FFFFFFFFH which gives zero
because of overflow in the INC instruction. If x = 0FFFFFFFFH is possible,
then change the code to:
MOV EDX,f
ADD EAX,1
JC DOVERFL
MUL EDX
DOVERFL:SHR EDX,b
If the value of x is limited, then you may use a lower value of r, i.e.
fewer digits. There can be several reasons to use a lower value of r:
- you may set r = 32 to avoid the SHR EDX,b in the end.
- you may set r = 16+b and use a multiplication instruction that gives
a 32 bit result rather than 64 bits. This will free the EDX register:
IMUL EAX,0CCCDh / SHR EAX,18
- you may choose a value of r that gives case C rather than case B in
order to avoid the INC EAX instruction
The maximum value for x in these cases is at least 2^(r-b), sometimes
higher. You have to do a systematic test if you want to know the exact
maximum value of x for which your code works correctly.
You may want to replace the slow multiplication instruction with faster
instructions as explained in paragraph 17.9.
The following example divides EAX by 10 and returns the result in EAX.
I have chosen r=17 rather than 19 because it happens to give a code,
which is easier to optimize, and covers the same range for x.
f = 2^17 / 10 = 3333h, case B: q = (x+1)*3333h:
LEA EBX,[EAX+2*EAX+3]
LEA ECX,[EAX+2*EAX+3]
SHL EBX,4
MOV EAX,ECX
SHL ECX,8
ADD EAX,EBX
SHL EBX,8
ADD EAX,ECX
ADD EAX,EBX
SHR EAX,17
A systematic test shows that this code works correctly for all x < 10004H.
Repeated division by the same value
-----------------------------------
If the divisor is not known at assembly time, but you are dividing
repeatedly with the same divisor, then you may use the same method as
above. The code has to distinguish between case A, B and C and calculate
f before doing the divisions.
The code that follows shows how to do many divisions with the same
divisor (unsigned division with truncation). First call SET_DIVISOR
to specify the divisor and calculate the reciprocal, then call
DIVIDE_FIXED for each value to divide by the same divisor.
.data
RECIPROCAL_DIVISOR DD ? ; rounded reciprocal divisor
CORRECTION DD ? ; case A: -1, case B: 1, case C: 0
BSHIFT DD ? ; number of bits in divisor - 1
.code
SET_DIVISOR PROC NEAR ; divisor in EAX
PUSH EBX
MOV EBX,EAX
BSR ECX,EAX ; b = number of bits in divisor - 1
MOV EDX,1
JZ ERROR ; error: divisor is zero
SHL EDX,CL ; 2^b
MOV [BSHIFT],ECX ; save b
CMP EAX,EDX
MOV EAX,0
JE SHORT CASE_A ; divisor is a power of 2
DIV EBX ; 2^(32+b) / d
SHR EBX,1 ; divisor / 2
XOR ECX,ECX
CMP EDX,EBX ; compare remainder with divisor/2
SETBE CL ; 1 if case B
MOV [CORRECTION],ECX ; correction for rounding errors
XOR ECX,1
ADD EAX,ECX ; add 1 if case C
MOV [RECIPROCAL_DIVISOR],EAX ; rounded reciprocal divisor
POP EBX
RET
CASE_A: MOV [CORRECTION],-1 ; remember that we have case A
POP EBX
RET
SET_DIVISOR ENDP
DIVIDE_FIXED PROC NEAR ; dividend in EAX, result in EAX
MOV EDX,[CORRECTION]
MOV ECX,[BSHIFT]
TEST EDX,EDX
JS SHORT DSHIFT ; divisor is power of 2
ADD EAX,EDX ; correct for rounding error
JC SHORT DOVERFL ; correct for overflow
MUL [RECIPROCAL_DIVISOR] ; multiply with reciprocal divisor
MOV EAX,EDX
DSHIFT: SHR EAX,CL ; adjust for number of bits
RET
DOVERFL:MOV EAX,[RECIPROCAL_DIVISOR] ; dividend = 0FFFFFFFFH
SHR EAX,CL ; do division by shifting
RET
DIVIDE_FIXED ENDP
This code gives the same result as the DIV instruction for
0 <= x < 2^32, 0 < d < 2^32.
Note: The line JC DOVERFL and its target are not needed if you are
certain that x < 0FFFFFFFFH.
If powers of 2 occur so seldom that it is not worth optimizing for them,
then you may leave out the jump to DSHIFT and in stead do a multiplication
with CORRECTION = 0 for case A.
If the divisor changes so often that SET_DIVISOR needs optimizing, then
you may replace the BSR instruction with the code given in section 17.7.
floating point division
-----------------------
Floating point division takes 39 clock cycles for the highest precision.
You can save time by specifying a lower precision in the floating point
control word (only FDIV and FIDIV are faster at low precision, no other
instructions can be speeded up this way).
It is possible to do a floating point division and an integer division in
parallel to save time. Example: A = A1 / A2; B = B1 / B2
FILD [B1]
FILD [B2]
MOV EAX,[A1]
MOV EBX,[A2]
CDQ
FDIV
DIV EBX
FISTP [B]
MOV [A],EAX
(make sure you set the floating point control word to the desired
rounding method)
Obviously, you should always try to minimize the number of divisions.
Floating point division with a constant or repeated division with the same
value should of course by done by multiplying with the reciprocal.
But there are many other situations where you can reduce the number of
divisions. For example:
if (A/B > C)... can be rewritten as if (A > B*C)... when B is
positive, and the opposite when B is negative.
A/B + C/D can be rewritten as (A*D + C*B) / (B*D)
If you are using integer division, then you should be aware that the
rounding errors may be different when you rewrite the formulas.
17.11 WAIT
----------
You can often increase speed by omitting the WAIT instruction.
The WAIT instruction has three functions:
a. The old 8087 processor requires a WAIT before _every_ floating point
instruction to make sure the coprocessor is ready to receive it.
b. WAIT is used to coordinate memory access between the floating point
unit and the integer unit. Examples:
b.1. FISTP [mem32]
WAIT ; wait for floating point unit to write before..
MOV EAX,[mem32] ; reading the result with the integer unit
b.2. FILD [mem32]
WAIT ; wait for floating point unit to read value..
MOV [mem32],EAX ; before overwriting it with integer unit
b.3. FLD QWORD PTR [ESP]
WAIT ; prevent an accidental hardware interrupt from..
ADD ESP,8 ; overwriting value on stack before it is read
c. WAIT is sometimes used to check for exceptions. It will generate an
interrupt if an unmasked exception bit in the floating point status
word has been set by a preceding floating point instruction.
Regarding a:
The function in point a is never needed on any other processors than the old
8087. Unless you want your code to be compatible with the 8087 you should
tell your assembler not to put in these WAITs by specifying a higher
processor. A 8087 floating point emulator also inserts WAIT instructions.
You should therefore tell your assembler not to generate emulation code
unless you need it.
Regarding b:
WAIT instructions to coordinate memory access are definitely needed on the
8087 and 80287 but not on the Pentium. It is not quite clear whether it is
needed on the 80387 and 80486. I have made several tests on these Intel
processors and not been able to provoke any error by omitting the WAIT on
any 32 bit Intel processor, although Intel manuals say that the WAIT is
needed for this purpose except after FNSTSW and FNSTCW. Omitting WAIT
instructions for coordinating memory access is not 100 % safe, even when
writing 32 bit code, because the code may be able to run on a 80386 main
processor with a 287 coprocessor, which requires the WAIT. Also, I have not
tested all possible hardware and software combinations, so there may be
other situations where the WAIT is needed.
If you want to be certain that your code will work on _any_ 32 bit processor
(including non-Intel processors) then I would recommend that you include the
WAIT here in order to be safe.
Regarding c:
The assembler automatically inserts a WAIT for this purpose before the
following instructions: FCLEX, FINIT, FSAVE, FSTCW, FSTENV, FSTSW.
You can omit the WAIT by writing FNCLEX, etc. My tests show that the WAIT
is unneccessary in most cases because these instructions without WAIT will
still generate an interrupt on exceptions except for FNCLEX and FNINIT on
the 80387. (There is some inconsistency about whether the IRET from the
interrupt points to the FN.. instruction or to the next instruction).
Almost all other floating point instructions will also generate an interrupt
if a previous floating point instruction has set an unmasked exception bit,
so the exception is likely to be detected sooner or later anyway. You may
insert a WAIT after the last floating point instruction in your program to
be sure to catch all exceptions.
You may still need the WAIT if you want to know exactly where an exception
occurred in order to be able to recover from the situation. Consider, for
example, the code under b.3 above: If you want to be able to recover from
an exception generated by the FLD here, then you need the WAIT because an
interrupt after ADD ESP,8 would overwrite the value to load.
17.12 FCOM + FSTSW AX
---------------------
The Pentium Pro and Pentium II processors have FCOMI instructions to
avoid the slow FSTSW. On processors without FCOMI instructions, the
usual way of doing floating point comparisons is:
FLD [a]
FCOMP [b]
FSTSW AX
SAHF
JB ASmallerThanB
You may improve this code by using FNSTSW AX rather than FSTSW AX and
test AH directly rather than using the non-pairable SAHF.
(TASM version 3.0 has a bug with the FNSTSW AX instruction)
FLD [a]
FCOMP [b]
FNSTSW AX
SHR AH,1
JC ASmallerThanB
Testing for zero or equality:
FTST
FNSTSW AX
AND AH,40H
JNZ IsZero ; (the zero flag is inverted!)
Test if greater:
FLD [a]
FCOMP [b]
FNSTSW AX
AND AH,41H
JZ AGreaterThanB
Do not use TEST AH,41H as it is not pairable. Do not use TEST EAX,4100H as
it would produce a partial register stall on the PentiumPro. Do not test the
flags after multibit shifts, as this has a penalty on the PentiumPro.
The FNSTSW instruction delays for 4 clocks after any floating point
instruction (probably because it is waiting for the status word to retire
from the pipeline). If you can fill this latency with integer instructions
between FCOM and FNSTSW, then the FNSTSW instruction will take only 2
clocks, rather than 6. Make sure you use the version without WAIT prefix
in this case.
It is sometimes faster to use integer instructions for comparing floating
point values, as described in paragraph 18 below.
17.13 freeing floating point registers
--------------------------------------
You have to free all used floating point registers before exiting a
subroutine, except for any register used for the result.
The fastest way to free one register is FSTP ST.
The fastest way to free two registers is FCOMPP.
It is not recommended to use FFREE.
17.14 FISTP
-----------
Converting a floating point number to integer is normally done like this:
FISTP DWORD PTR [TEMP]
MOV EAX, [TEMP]
An alternative method is:
.DATA
ALIGN 8
TEMP DQ ?
MAGIC DD 59C00000H ; f.p. representation of 2^51 + 2^52
.CODE
FADD [MAGIC]
FSTP QWORD PTR [TEMP]
MOV EAX, DWORD PTR [TEMP]
Adding the 'magic number' of 2^51 + 2^52 has the effect that any integer
between -2^31 and +2^31 will be aligned in the lower 32 bits when storing
as a double precision floating point number. The result is the same as you
get with FISTP for all rounding methods except truncation towards zero.
The result is different from FISTP if the control word specifies truncation
or in case of overflow. You may need a WAIT instruction for compatibility
with older processors, see section 17.11.
This method is not faster than using FISTP, but it gives better scheduling
opportunities because there is a 3 clock void between FADD and FSTP which
may be filled with other instrucions. You may multiply or divide the number
by a power of 2 in the same operation by doing the opposite to the magic
number. You may also add a constant by adding it to the magic number,
which then has to be double precision.
17.15 FPTAN
-----------
According to the manuals, FPTAN returns two values X and Y and leaves it to
the programmer to divide Y with X to get the result, but in fact it always
returns 1 in X so you can save the division. My tests show that on all 32
bit Intel processors with floating point unit or coprocessor, FPTAN always
returns 1 in X regardless of the argument. If you want to be absolutely sure
that your code will run correctly on all processors, then you may test if X
is 1, which is faster than dividing with X. The Y value may be very high,
but never infinity, so you don't have to test if Y contains a valid number
if you know that the argument is valid.
18. USING INTEGER INSTRUCTIONS TO DO FLOATING POINT OPERATIONS
==============================================================
Integer instructions are generally faster than floating point instructions,
so it is often advantageous to use integer instructions for doing simple
floating point operations. The most obvious example is moving data. Example:
FLD QWORD PTR [ESI] / FSTP QWORD PTR [EDI]
Change to:
MOV EAX,[ESI] / MOV EBX,[ESI+4] / MOV [EDI],EAX / MOV [EDI+4],EBX
The former code takes 4 clocks, the latter takes 2.
Testing if a floating point value is zero:
The floating point value of zero is usually represented as 32 or 64 bits of
zero, but there is a pitfall here: The sign bit may be set! Minus zero is
regarded as a valid floating point number, and the processor may actually
generate a zero with the sign bit set if for example multiplying a negative
number with zero. So if you want to test if a floating point number is zero,
you should not test the sign bit. Example:
FLD DWORD PTR [EBX] / FTST / FNSTSW AX / AND AH,40H / JNZ IsZero
Use integer instructions instead, and shift out the sign bit:
MOV EAX,[EBX] / ADD EAX,EAX / JZ IsZero
The former code takes 9 clocks, the latter takes only 2.
If the floating point number is double precision (QWORD) then you only have
to test bit 32-62. If they are zero, then the lower half will also be zero
if it is a normal floating point number.
Testing if negative:
A floating point number is negative if the sign bit is set and at least one
other bit is set. Example:
MOV EAX,[NumberToTest] / CMP EAX,80000000H / JA IsNegative
Manipulating the sign bit:
You can change the sign of a floating point number simply by flipping the
sign bit. Example:
XOR BYTE PTR [a] + (TYPE a) - 1, 80H
Likewise you may get the absolute value of a floating point number by simply
ANDing out the sign bit.
Comparing numbers:
Floating point numbers are stored in a unique format which allows you to use
integer instructions for comparing floating point numbers, except for the
sign bit. If you are certain that two floating point numbers both are normal
and positive then you may simply compare them as integers. Example:
FLD [a] / FCOMP [b] / FNSTSW AX / AND AH,1 / JNZ ASmallerThanB
Change to:
MOV EAX,[a] / MOV EBX,[b] / CMP EAX,EBX / JB ASmallerThanB
This method only works if the two numbers have the same precision and you
are certain that none of the numbers have the sign bit set.
If negative numbers are possible, then you have to convert the
negative numbers to 2-complement, and do a signed compare:
MOV EAX,[a]
MOV EBX,[b]
MOV ECX,EAX
MOV EDX,EBX
SAR ECX,31 ; copy sign bit
AND EAX,7FFFFFFFH ; remove sign bit
SAR EDX,31
AND EBX,7FFFFFFFH
XOR EAX,ECX ; make 2-complement if sign bit was set
XOR EBX,EDX
SUB EAX,ECX
SUB EBX,EDX
CMP EAX,EBX
JL ASmallerThanB ; signed comparison
This method works for all normal floating point numbers, including -0.
19. USING FLOATING POINT INSTRUCTIONS TO DO INTEGER OPERATIONS
==============================================================
19.1 Moving data
----------------
Floating point instructions can be used to move 8 bytes at a time:
FILD QWORD PTR [ESI] / FISTP QWORD PTR [EDI]
This is only an advantage if the destination is not in the cache. The
optimal way to move a block of data to uncached memory on the Pentium is:
TopOfLoop:
FILD QWORD PTR [ESI]
FILD QWORD PTR [ESI+8]
FXCH
FISTP QWORD PTR [EDI]
FISTP QWORD PTR [EDI+8]
ADD ESI,16
ADD EDI,16
DEC ECX
JNZ TopOfLoop
The source and destination should of course be aligned by 8. The extra time
used by the slow FILD and FISTP instructions is compensated for by the fact
that you only have to do half as many write operations. Note that this
method is only advantageous on the Pentium and only if the destination is
not in the level 1 cache. On all other processors the optimal way to move
blocks of data is REP MOVSD, or if you have a processor with MMX you may use
the MMX instructions instead to write 8 bytes at a time.
19.2 Integer multiplication
---------------------------
Floating point multiplication is faster than integer multiplication on the
Pentium without MMX, but the price for converting integer factors to float
and converting the result back to integer is high, so floating point
multiplication is only advantageous if the number of conversions needed is
low compared with the number of multiplications. Integer multiplication is
faster than floating point on other processors.
It may be tempting to use denormal floating point operands to save some of
the conversions here, but the handling of denormals is very slow, so this
is not a good idea!
19.3 Integer division
---------------------
Floating point division is not faster than integer division, but you can do
other integer operations (including integer division, but not integer
multiplication) while the floating point unit is working on the division.
See paragraph 17.10 above for an example.
19.4 Converting binary to decimal numbers
-----------------------------------------
Using the FBSTP instruction is a simple and convenient way of converting a
binary number to decimal, although not necessarily the fastest method.
20. LIST OF INTEGER INSTRUCTIONS
================================
Explanations:
Operands: r=register, m=memory, i=immediate data, sr=segment register
m32= 32 bit memory operand, etc.
Clock cycles:
The numbers are minimum values. Cache misses, misalignment, and exceptions
may increase the clock counts considerably.
Pairability:
u=pairable in U-pipe, v=pairable in V-pipe, uv=pairable in either pipe,
np=not pairable
Opcode Operands Clock cycles Pairability
----------------------------------------------------------------------------
NOP 1 uv
MOV r/m, r/m/i 1 uv
MOV r/m, sr 1 np
MOV sr , r/m >= 2 b) np
MOV m , accum 1 uv h)
XCHG (E)AX, r 2 np
XCHG r , r 3 np
XCHG r , m >20 np
XLAT 4 np
PUSH r/i 1 uv
POP r 1 uv
PUSH m 2 np
POP m 3 np
PUSH sr 1 b) np
POP sr >= 3 b) np
PUSHF 4 np
POPF 6 np
PUSHA POPA 5 np
LAHF SAHF 2 np
MOVSX MOVZX r, r/m 3 a) np
LEA r/m 1 uv
LDS LES LFS LGS LSS m 4 c) np
ADD SUB AND OR XOR r , r/i 1 uv
ADD SUB AND OR XOR r , m 2 uv
ADD SUB AND OR XOR m , r/i 3 uv
ADC SBB r , r/i 1 u
ADC SBB r , m 2 u
ADC SBB m , r/i 3 u
CMP r , r/i 1 uv
CMP m , r/i 2 uv
TEST r , r 1 uv
TEST m , r 2 uv
TEST r , i 1 f)
TEST m , i 2 np
INC DEC r 1 uv
INC DEC m 3 uv
NEG NOT r/m 1/3 np
MUL IMUL r8/r16/m8/m16 11 np
MUL IMUL all other versions 9 d) np
DIV r8/m8 17 np
DIV r16/m16 25 np
DIV r32/m32 41 np
IDIV r8/m8 22 np
IDIV r16/m16 30 np
IDIV r32/m32 46 np
CBW CWDE 3 np
CWD CDQ 2 np
SHR SHL SAR SAL r , i 1 u
SHR SHL SAR SAL m , i 3 u
SHR SHL SAR SAL r/m, CL 4/5 np
ROR ROL RCR RCL r/m, 1 1/3 u
ROR ROL r/m, i(><1) 1/3 np
ROR ROL r/m, CL 4/5 np
RCR RCL r/m, i(><1) 8/10 np
RCR RCL r/m, CL 7/9 np
SHLD SHRD r, i/CL 4 a) np
SHLD SHRD m, i/CL 5 a) np
BT r, r/i 4 a) np
BT m, i 4 a) np
BT m, r 9 a) np
BTR BTS BTC r, r/i 7 a) np
BTR BTS BTC m, i 8 a) np
BTR BTS BTC m, r 14 a) np
BSF BSR r , r/m 7-73 a) np
SETcc r/m 1/2 a) np
JMP CALL short/near 1/4 e) v
JMP CALL far >= 3 e) np
conditional jump short/near 1/4/5 e) v
CALL JMP r/m 2/5 e) np
RETN 2/5 e) np
RETN i 3/6 e) np
RETF 4/7 e) np
RETF i 5/8 e) np
J(E)CXZ short 5-8 e) np
LOOP short 5-9 e) np
BOUND r , m 8 np
CLC STC CMC CLD STD 2 np
CLI STI 6-7 np
LODS 2 np
REP LODS 7+3*n g) np
STOS 3 np
REP STOS 10+n g) np
MOVS 4 np
REP MOVS 12+n g) np
SCAS 4 np
REP(N)E SCAS 9+4*n g) np
CMPS 5 np
REP(N)E CMPS 8+4*n g) np
BSWAP 1 a) np
CPUID 13/15/16 a) np
RDTSC 6 a) np
----------------------------------------------------------------------------
Notes:
a) this instruction has a 0FH prefix which takes one clock cycle extra to
decode on a Pentium without MMX unless preceded by a multicycle
instruction (see section 13 above).
b) versions with FS and GS have a 0FH prefix. see note a.
c) versions with SS, FS, and GS have a 0FH prefix. see note a.
d) versions with two operands and no immediate have a 0FH prefix, see note a.
e) see section 12 above
f) only pairable if register is accumulator. see paragraph 17.3 above
g) add one clock cycle for decoding the repeat prefix unless preceded by a
multicycle instruction (such as CLD. see section 13 above).
h) pairs as if it were writing to the accumulator. see paragraph 17.2 above.
21. LIST OF FLOATING POINT INSTRUCTIONS
=======================================
Explanations:
Operands: r=register, m=memory, m32=32 bit memory operand, etc.
Clock cycles:
The numbers are minimum values. Cache misses, misalignment, denormal
operands, and exceptions may increase the clock counts considerably.
Pairability:
+=pairable with FXCH, np=not pairable with FXCH.
i-ov:
Overlap with integer instructions. i-ov = 4 means that the last four clock
cycles can overlap with subsequent integer instructions.
fp-ov:
Overlap with floating point instructions. fp-ov = 2 means that the last two
clock cycles can overlap with subsequent floating point instructions.
(WAIT is considered a floating point instruction here)
Opcode Operand Clock cycles Pairability i-ov fp-ov
-----------------------------------------------------------------------------
FLD r/m32/m64 1 + 0 0
FLD m80 3 np 0 0
FBLD m80 48-58 np 0 0
FST(P) r 1 np 0 0
FST(P) m32/m64 2 i) np 0 0
FST(P) m80 3 i) np 0 0
FBSTP m80 148-154 np 0 0
FILD m 3 np 2 2
FIST(P) m 6 np 0 0
FLDZ FLD1 2 np 0 0
FLDPI FLDL2E etc. 5 np 0 0
FNSTSW AX/m16 6 n) np 0 0
FLDCW m16 8 np 0 0
FNSTCW m16 2 np 0 0
FADD(P) r/m 3 + 2 2
FSUB(R)(P) r/m 3 + 2 2
FMUL(P) r/m 3 + 2 2 j)
FDIV(R)(P) r/m 19/33/39 m) + k) 38 l) 2
FCHS FABS 1 + 0 0
FCOM(P)(P) FUCOM r/m 1 + 0 0
FIADD FISUB(R) m 6 np 2 2
FIMUL m 6 np 2 2
FIDIV(R) m 22/36/42 m) np 38 l) 2
FICOM m 4 np 0 0
FTST 1 np 0 0
FXAM 17-21 np 4 0
FPREM 16-64 np 2 2
FPREM1 20-70 np 2 2
FRNDINT 9-20 np 0 0
FSCALE 20-32 np 5 0
FXTRACT 12-66 np 0 0
FSQRT 70 np 69 l) 2
FSIN FCOS 16-126 np 2 2
FSINCOS 17-137 np 2 2
F2XM1 13-57 np 2 2
FYL2X 22-111 np 2 2
FYL2XP1 22-103 np 2 2
FPATAN 19-134 np 2 2
FPTAN 17-173 np 36 l) 0
FNOP 2 np 0 0
FXCH r 1 np 0 0
FINCSTP FDECSTP 2 np 0 0
FFREE r 2 np 0 0
FNCLEX 6-9 np 0 0
FNINIT 12-22 np 0 0
FNSAVE m 124-300 np 0 0
FRSTOR m 70-95 np 0 0
WAIT 1 np 0 0
-----------------------------------------------------------------------------
Notes:
i) The value to store is needed one clock cycle in advance.
j) 1 if the overlapping instruction is also an FMUL.
k) If the FXCH is followed by an integer instruction then it will pair
imperfectly and take an extra clock cycle so that the integer
instruction will begin in clock cycle 3.
l) Cannot overlap integer multiplication instructions.
m) FDIV takes 19, 33, or 39 clock cycles for 24, 53, and 64 bit precision
respectively. FIDIV takes 3 clocks more. The precision is defined by bit
8-9 of the floating point control word.
n) The first 4 clock cycles can overlap with preceding integer instructions.
See section 17.12 above.
22. MMX instructions
====================
A list of MMX instruction timings is not needed because they all take
one clock cycle, except the MMX multiply instructions which take 3.
MMX multiply instructions can be overlapped and pipelined to yield a
throughput of one multiplication per clock cycle.
The EMMS instruction takes only one clock cycle, but the first floating
point instruction after an EMMS takes 56 clocks extra, and the first
MMX instruction after a floating point instruction takes 38 clocks extra
and is non-pairable. There is no penalty for an MMX instruction after EMMS.
There is no penalty for using a memory operand in an MMX instruction
because the MMX arithmetic unit is one step later in the pipeline than
the load unit. But the penalty comes when you store data from an MMX
register to memory or to a 32 bit register: The data have to be ready
two clock cycles in advance. This is analogous to the floating point
store instructions.
All MMX instructions except EMMS are pairable in either pipe. Pairing
rules for MMX instructions are described in paragraph 8.10 above.
23. TESTING SPEED
=================
The Pentium has an internal 64 bit clock counter which can be read into
EDX:EAX using the instruction RDTSC (read time stamp counter). This is very
useful for testing exactly how many clock cycles a piece of code takes.
The program below is useful for measuring the number of clock cycles a piece
of code takes. The program executes the code to test 10 times and stores the
10 clock counts. The program can be used in both 16 and 32 bit mode.
RDTSC MACRO ; define RDTSC instruction
DB 0FH,31H
ENDM
ITER EQU 10 ; number of iterations
.DATA ; data segment
ALIGN 4
COUNTER DD 0 ; loop counter
TICS DD 0 ; temporary storage of clock
RESULTLIST DD ITER DUP (0) ; list of test results
.CODE ; code segment
BEGIN: MOV [COUNTER],0 ; reset loop counter
TESTLOOP: ; test loop
;**************** Do any initializations here: ************************
FINIT
;**************** End of initializations ************************
RDTSC ; read clock counter
MOV [TICS],EAX ; save count
CLD ; non-pairable filler
REPT 8
NOP ; eight NOP's to avoid shadowing effect
ENDM
;**************** Put instructions to test here: ************************
FLDPI ; this is only an example
FSQRT
RCR EBX,10
FSTP ST
;********************* End of instructions to test ************************
CLC ; non-pairable filler with shadow
RDTSC ; read counter again
SUB EAX,[TICS] ; compute difference
SUB EAX,15 ; subtract the clock cycles used by fillers etc
MOV EDX,[COUNTER] ; loop counter
MOV [RESULTLIST][EDX],EAX ; store result in table
ADD EDX,TYPE RESULTLIST ; increment counter
MOV [COUNTER],EDX ; store counter
CMP EDX,ITER * (TYPE RESULTLIST)
JB TESTLOOP ; repeat ITER times
; insert here code to read out the values in RESULTLIST
The 'filler' instructions before and after the piece of code to test are
critical. The CLD is a non-pairable instruction which has been inserted to
make sure the pairing is the same the first time as the subsequent times.
The eight NOP instructions are inserted to prevent any prefixes in the code
to test to be decoded in the shadow of the preceding instructions. Single
byte instructions are used here to obtain the same pairing the first time as
the subsequent times. The CLC after the code to test is a non-pairable
instruction which has a shadow under which the 0FH prefix of the RDTSC can
be decoded so that it is independent of any shadowing effect from the code
to test.
On the PentiumPro you have to put in a serializing instruction like CPUID
before and after each RDTSC to prevent it from executing in parallel with
anything else. CPUID has a decoding delay on the Pentium without MMX because
of the 0FH prefix, but it has no shadow because it is serializing.
The RDTSC instruction cannot execute in virtual mode on the Pentium
without MMX, so if you are running under DOS you must skip the EMM386
(or any other memory manager) in your CONFIG.SYS and not run under a DOS
box in Windows. If you have Windows 95 you should press F8 while booting
and choose 'safe mode command prompt only'.
The Pentium processors have special performance monitor counters which can
count events such as cache misses, misalignments, AGI stalls, etc. Details
about how to use the performance monitor counters are not covered by this
manual but can be found in the MMX technology developer's manual.
24. PENTIUM PRO AND PENTIUM II MICROPROCESSORS
==============================================
The Pentium Pro and Pentium II processors are very different from the
earlier versions. Each instruction is decoded into micro-operations
(uops) which are reordered before execution. The main advantage of this
is that the processor does most of the optimations for you: splitting
complex instructions into simpler ones, and reordering instructions so
that independent operations can be done in parallel. These dynamic
processors therefore perform better on non-optimized code than before.
Another advantage is automatic register renaming. Example:
MOV EAX,[ESI]
INC EAX
MOV [EDI],EAX
MOV EAX,[ESI+4]
INC EAX
MOV [EDI+4],EAX
In this case the dynamic processor will use a different physical register
for the last three instructions and rename it to EAX so that the two
calculations can be done in parallel. In case of a cache miss for [ESI]
and a hit for [ESI+4], you will actually have the last calculation done
first. There are more physical registers than logical register names, so
you can take advantage of this automatic register renaming if you are
short on registers.
The dynamic processors also have drawbacks, however. Most importantly,
mispredicted jumps are very expensive (10-16 clock cycles!). To
compensate for this problem, the dynamic processors have conditional
move instructions which you can use in some situations in stead of
conditional jumps.
Another drawback is partial register stalls and partial memory stalls.
Using a full register after writing to part of the register will cause a
severe delay on the PentiumPro and PentiumII. Example:
MOV AL,[EBX] / MOV ECX,EAX
You may avoid this penalty by zeroing the full register first:
XOR EAX,EAX / MOV AL,[EBX] / MOV ECX,EAX
or by using MOVZX EAX,BYTE PTR [EBX]
A similar penalty occurs when you address the same piece of memory with
different data sizes. Example:
MOV [ESI],EAX / MOV BL,[ESI]
A thorough optimation for the dynamic processors require that you know
how ASM instructions are broken down into uops. You will need Intel's
application note: 'AP-526 Optimizations for Intel's 32-bit Processors'
which has a table of this in appendix D. The tutorial 'Optimizing for
the Pentium® Pro Processor Family' is also recommended.
The maximum throughput in the PentiumPro and Pentium II processors is
3 uops per clock cycle. In order to achieve this, you should beware of
the possible bottlenecks. The instruction decode stage can decode up to
three ASM instructions per clock cycle, but only if the first of these
generates no more than 4 uops and the two next instructions generate
only 1 uop each. This means that a throughput of 6 uops in the decode
stage is achieved with a 4-1-1-4-1-1 sequence, whereas you get only 2
uops per clock if all instructions generate 2 uops each. In order to get
a throughput in the decode stage of at least 3 uops per clock cycle you
should, if possible, intersperce instructions that generate more than
one uop with one or two instructions that generate only one uop each.
More often you will get a bottleneck in the execution stage. There are
five execution pipelines called port 0-4. Port 0 and 1 correspond roughly
to the u and v pipes of the Pentium. Port 0 can do all artihmetic
operations whereas port 1 can do only simple ALU operations. Port 2 does
all memory read operations. A memory write operation requires two uops
which go to port 3 and 4 respectively.
Each port can receive only one uop per clock cycle. In order to get the
maximum throughput of 3 uops in the execution stage you must make sure
that no port gets more than one third of the uops. If, for example, you
have too many uops for port 0 and 1, then you may replace some
MOV reg,reg or MOV reg,constant instructions with MOV reg,[mem]
instructions, which go to port 2 instead or port 0 or 1.
You also have to take data-flow dependency into account. If all uops
depend on the result of the preceding one, then you can execute only one
per clock cycle. The dynamic processor can reorder your uops in order to
first execute any operation that does not depend on the preceding ones,
but this requires that your code is not completely sequential.
Normally, you wouldn't know what the reorder-buffer contains at the
start of the piece of code you are optimizing, but if you are coding a
loop you can assume that you will reach a steady-state after one or a
few iterations, so that you only have to count the uops inside the loop.
25. CONSIDERATIONS FOR OTHER MICROPROCESSORS
============================================
The following table summarizes some important differences between the
microprocessors in the Pentium family:
Pentium Pentium Pentium Pentium
MMX Pro II
--------------------------------------------------------------------
code cache, kb 8 16 8 16
data cache, kb 8 16 8 16
MMX instructions no yes no yes
conditional move instructructions no no yes yes
dynamic execution no no yes yes
branch prediction poor good good good
return stack buffer no yes yes yes
branch misprediction penalty 3-4 4-5 10-16 10-16
partial register stall 0 0 >6 ?
FMUL latency 3 3 5 5
FMUL throughput 1/2 1/2 1/2 1/2
IMUL latency 9 9 4-5 4-5
IMUL throughput 1/9 1/9 1/1 1/1
--------------------------------------------------------------------
Comments to the table:
Code cache size is important if the critical part of your program is
not limited to a small memory space.
Data cache size is important for all programs that handle more than
small amounts of data in the critical part.
MMX instructions are useful for programs that handle massively parallel
integer data, such as sound and image processing. In other applications
it may not be possible to take advantage of the MMX instructions.
Conditional move instructructions are useful for avoiding poorly
predictable conditional jumps.
Dynamic execution improves performance on non-optimized code. It includes
automatic instruction reordering and register renaming.
Processors with a good branch prediction method can predict simple
repetitive patterns, such as a branch instruction which falls through
every fourth time. A good branch prediction is most important if the
branch misprediction penalty is high.
A return stack buffer improves prediction of return instructions when
a subroutine is called alternatingly from different locations.
A partial register stall makes handling of mixed data sizes (8, 16, 32 bit)
more difficult.
The latency of a multiplication instruction is the time it takes in
sequential code. A throughput of 1/2 means that the execution can be
pipelined so that a new multiplication can begin every second clock cycle.
This defines the speed for handling parallel data.
As you can see from the table, the Pentium II processor will be the
best choise in most situations, but definitely not in all situations.
A fully optimized program with a lot of sequential floating point code
and a lot of unpredictable branches might actually run fastest on the
simple Pentium without MMX (at the same clock frequency).
Most of the optimations described in this document have little or no
negative effects on other microprocessors, including non-Intel processors,
but there are some problems to be aware of.
Scheduling floating point code for the Pentium often requires a lot of
extra FXCH instructions. This will slow down execution on earlier
microprocessors, but not on the Pentium family and advanced non-Intel
processors.
The MMX instructions available on the Pentium MMX and Pentium II processors
are very useful for massively parallel integer calculations, but they
cannot be used without a serious penalty if you have floating point
instructions in the same part of the program.
Taking advantage of the new instructions in the MMX and Pentium Pro
processors will create problems if you want your code to be compatible
with earlier microprocessors. The solution may be to write several
versions of your code, each optimized for a particular processor. Your
program should automatically detect which processor it is running on and
select the appropriate version of code. Such a complicated approach is of
course only needed for the most critical parts of your program.