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HOW TO OPTIMIZE FOR THE PENTIUM PROCESSOR
Copyright (c) 1996 by Agner Fog
4. memory model
7. pairing integer instructions
8. first time vs. repeated execution
9. pairing multicycle instructions
10. splitting complex instructions into simpler ones
11. address generation interlock
12. jumps and branches
14. reducing code size
15. scheduling floating point code
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. testing code speed
23. considerations for other microprocessors
This manual describes in detail how to write optimized code for the
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 already is 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
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
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.
Some high level language compilers offer reasonably 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.
Intel has recently announced that they will produce new versions of the
Pentium and PentiumPro processors with 57 new instruction opcodes for doing
integer vector operations. These are called multimedia extension (MMX)
instructions. The Pentium chip with MMX will behave slightly different from
the Pentium without MMX, according to the technical documents. These
differences will be mentioned where appropriate. The Pentium Pro processor
behaves very differently from the Pentium, and will be covered only briefly
by this manual.
The informations in this document are based mainly on my own experiments on
a Pentium computer. Informations about PentiumPro and MMX processors are
based solely on documents from Intel.
Good luck with your hunt for nanoseconds!
A lot of useful literature can be downloaded for free from Intel's www site:
You can find the documents you need by using the search facilities at:
The documents are in various different file formats. If a particular
document is in a format not supported by your word processing software then
you may seek an appropriate file viewer somewhere on the internet. Many
software companies are offering such file viewers for free to support their
The most useful document is Intel's application note:
'AP-526 Optimizations for Intel's 32-bit Processors'
which can be downloaded from the following URL:
A fancy tutorial named "Optimizing Applications for the Pentium Processor"
is awailable at:
Detailed information on the MMX processors can be found in the documents:
"Intel architecture MMX technology developer's manual", and
"Intel architecture MMX technology programmers reference manual"
both of which are awailable from:
A lot of other sources than Intel also have useful information. These sources
are listed in the comp.lang.asm.x86 FAQ. I would particularly recommend
The shareware editor ASMEDIT has an online help which covers all instruction
codes etc. It is available from:
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).
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.
All data in RAM should be aligned to adresses divisible by 2, 4, or 8
according to this scheme:
operand size alignment
1 (byte) 1
2 (word) 2 (or adress 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 8 or 16.
The Pentium has 8 kb of on-chip cache (level one cache) for code, and 8 kb
for data. Data in the level 1 cache can be read or written to in only 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 effectively. Most descriptions of the cache in other documents are
insufficient and difficult to understand. I hope you find this one better.
The data cache 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 aligned 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 which are used near each other into
aligned blocks of 32 bytes of memory. If, for example, you have a loop which
accesses two arrays, then you may combine 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 bit 5 through 11 of the physical
RAM address (bit 0-4 define the 32 bytes within a cache line). There are two
cache lines for each of the 128 set-values, so there are two possible cache
lines to assign to any RAM address. The consequence of this is that the
cache can hold no more than two 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]
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. 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 that was used for EAX,
and fills it with the data from [ESI + 20*4096] to [ESI + 20*4096 + 31] and
then reads ECX. Next, when reading EAX, you find that the cache line that
held the value for EAX has 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 of
max. 8 kbytes or two contiguous blocks of max. 4 kbytes each (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 main 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), 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 MB 72 pin RAM modules, and 2 kb for
16 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 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, which always loads 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 cache. The method is explained in
section 19 below.
Temporary data may conveniently be stored at the stack because stack data
are 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.)
The Pentium has a separate 8 kb cache for code, which is similar to the data
cache. 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 go to the bottom of your code.
7. 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
INC, DEC, ADD, SUB, CMP, AND, OR, XOR,
and some forms of TEST (see section 17.1)
The following instructions are pairable in the U-pipe only:
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
All other integer instructions can execute in the U-pipe only, and are not
Two consecutive instructions will pair when the following conditions are met:
7.1 The first instruction is pairable in the U-pipe and the second instruction
is pairable in the V-pipe.
7.2 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
7.3 In rule 7.2 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
7.4 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] ; within same dword, do not pair.
MOV AL, [ESI+3] / MOV BL, [ESI+4] ; on each side of a dword boundary, pair.
7.5 Rule 7.4 is extended to the case where bit 2-4 is the same in the two
adresses. For dword adresses this means that the difference between the
two adresses should not be divisible by 32. Examples:
MOV [ESI], EAX / MOV [ESI+32000], EBX ; do not pair
MOV [ESI], EAX / MOV [ESI+32004], EBX ; pair OK
7.6 Two instructions which both write to parts of the flags register can pair
despite rule 7.2 and 7.3. Example:
SHR EAX,4 / INC EBX ; pair OK
7.7 An instruction which writes to the flags can pair with a conditional jump
despite rule 7.2. Example:
CMP EAX, 2 / JA LabelBigger ; pair OK
7.8 The following instruction combinations can pair despite the fact that they
both modify the stack pointer:
PUSH + PUSH, PUSH + CALL, POP + POP
PUSH EAX / CALL MySub ; pair OK
PUSH AX / PUSH BX ; (16 bit mode) do not pair if SP is divisible by 4
; because of rule 7.4
7.9 There are restrictions on the pairing on instructions with prefix.
There are several types of prefixes:
- instructions adressing 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
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.
7.10 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:, 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
A further condition is that both instructions must be preloaded and decoded.
This is explained in the next section:
8. FIRST TIME VERSUS REPEATED EXECUTION
Loading and decoding instructions may take more time than executing them.
Therefore code which is not in the cache may be delayed by the time it takes
to read the code from RAM.
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.
For these two reasons, a piece of code inside a loop will generally take
much more time the first time it executes than the subsequent times. A third
reason is that any data used by this piece of code may not be cached the
If you have a big loop which doesn't fit in the code cache then you will get
the first-time penalty all the time because it doesn't run from the cache.
9. PAIRING MULTICYCLE INSTRUCTIONS
Pairable instructions which do not access memory take one clock cycle,
except 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 penalty for using complex addressing modes such as scaled index
Pairable instructions which read from memory, does some calculation, and
stores the result in a register or flags, take 2 clock cycles. (read/modify
Pairable instructions which read from memory, does some calculation, and
writes the result back to the memory, take 3 clock cycles.
The number of clock cycles used by two paired instructions is:
| 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
ADD [mem1], EAX / ADD EBX, [mem2] ; 4 clock cycles
ADD EBX, [mem2] / ADD [mem1], EAX ; 3 clock cycles
When two paired instructions both take extra time due to cache misses,
misalignment, or jump misprediction, then the pair takes more time than each
instruction, but less than the sum of the two.
10. SPLITTING COMPLEX INSTRUCTIONS INTO SIMPLER ONES
You may split up read/modify and read/modify/write instructions to improve
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
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 instruction in order to reduce code size.
11. 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.
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
adressing, 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.
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
12. JUMPS AND BRANCHES
The Pentium attempts to predict whether a conditional jump is taken or not.
It uses a 'branch target buffer' (BTB), which can remember the history of
256 jump instructions.
The Pentium without MMX makes the prediction on the basis of the last two
events. It guesses that a conditional jump will be taken if it was taken
the previous time or the time before. It guesses that the jump is not taken
if it was not taken the last two times. A conditional jump which has not
been seen before (or is not in the BTB) is predicted as not taken.
The Pentium with MMX (and the PentiumPro) makes its prediction on the basis
of the last four events, so that it is able to predict a simple repetitive
pattern. A conditional jump which has not been seen before (or is not in the
BTB) is predicted as taken if it goes backwards (e.g. a loop), and as not
taken if it goes forward.
If a conditional jump is properly predicted (i.e. if the guess was correct)
then it takes only one clock cycle. A mispredicted branch takes 4 clock
cycles if it is in the U-pipe and 5 clock cycles if it is in the V-pipe.
The penalty for misprediction is doubled or tripled if another jump or call
follows in the first clock cycle after the mispredicted branch. The first
two instructions after a possibly mispredicted branch should therefore not
be branch, jump, or call instructions. You may put in NOP's or other
instructions to avoid this.
The Pentium with MMX may also have a penalty in this case, but only if the
two consequtive branch instructions end within the same aligned dword of
memory. This problem may be avoided by using a near displacement rather
than short on the second branch instruction to make it longer, but this
method doesn't help on the Pentium without MMX so you should rather put in
some NOP's or other instructions.
The jump prediction algorithm is optimal for a loop where the testing is at
the bottom, as in this example:
MOV ECX, [N]
L: MOV [EDI],EAX
Since the jump prediction algorithm for the Pentium without MMX is
asymmetric, there may be situations where you can improve performance by
reordering your code. Consider for example this if-then-else construction:
JNZ SHORT A1
JMP SHORT E
A1: CALL F1
If F0 is called more often than F1, and F1 is seldom called twice in
succession, then you can improve jump prediction on the Pentium without MMX
by swapping the two branches. However, This will be slightly suboptimal on
the Pentium with MMX and the PentiumPro because they may mispredict the
branch if it is not in the branch target buffer. Another tradeoff is that
the code cache is used less efficiently when the seldom used branch comes
first. You may put in two NOP's before each of the CALL instructions here to
avoid the penalty of a call after a mispredicted jump.
Indirect jumps and calls through function pointers will be poorly predicted
on the Pentium with MMX and the PentiumPro.
All calls should be matched with returns in order for the returns to be
predicted correctly on the Pentium with MMX and the PentiumPro.
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:
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;
This example chooses between two numbers: if (a != 0) a = b; else a = c;
Whether or not such tricks are worth the extra code depend on how
predictable a conditional jump would be. Un a PentiumPro you can use
conditional move instructions to obtain the same effect.
An instruction with one or more prefixes may not be able to execute in the
V-pipe (se paragraph 7.9), and it takes one clock cycle extra for each
prefix to decode, except for 0FH prefixes on the Pentium with MMX and
conditional near jumps.
Adress 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. On the Pentium with MMX, unpaired instructions also have a shadowing
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
14. REDUCING CODE SIZE
As explained in paragraph 6, the code cache is 8 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 adresses 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 rule 7.4 and 7.8 above). Some
other methods for reducing the size or your code are discussed below.
Both jump adresses, data adresses, 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 adresses this means that short jumps take two bytes of code,
whereas jumps beyond 127 bytes take 5 bytes if unconditional and 6 bytes if
Likewise, data adresses take less space if they can be expressed as a
pointer and a displacement between -128 and +127.
MOV EBX,DS: / ADD EBX,DS: ; 12 bytes
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
adresses, 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
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 only 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
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: is smaller than MOV EBX,DS:
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.
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
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
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
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
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]
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
Paragraph 21 gives a complete listing of floating point instructions, and
what they can pair or overlap with.
One floating point instruction requires special mentioning, namely FST or
FSTP with a memory operand. This instruction takes two clock cycles in the
execution stage, but it seems to start converting the value in ST(0) already
at the address decode stage in the pipeline, so that there is 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. No other
instructions have this weirdness. The two clock cycles in the execution
stage of the FST(P) instruction cannot pair or overlap with any subsequent
Instructions with integer operands such as FIADD, FISUB, FIMUL, FIDIV, FICOM
may be split up into simpler operations in order to improve overlapping.
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.