Copyright (c) 1996 by Agner Fog


1.  introduction
2.  literature
3.  debugging
4.  memory model
5.  alignment
6.  cache
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
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. testing code speed
23. considerations for other microprocessors

This manual describes in detail how to write optimized code for the 
Pentium(R) microprocessor.

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 
file formats.  

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:\docs\microproc\general\app_note\242816+0

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).

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:

        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. 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.

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
  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 
conditional jump.

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 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, 
      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.

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:[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

A further condition is that both instructions must be preloaded and decoded.
This is explained in the next section:

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 
first time.

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.

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. 
(read/modify/write instructions).

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.

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 [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 instruction in order to reduce code size.

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:

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

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
        ADD EDI,4
        DEC ECX
        JNZ L

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: 

        TEST EAX,EAX
        JNZ  SHORT A1
        CALL F0
        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.

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:

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 


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.

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 

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:[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 
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:[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. 
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].  

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, 

- 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 
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 

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.  

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.

Discuss this article in the forums

Date this article was posted to 7/16/1999
(Note that this date does not necessarily correspond to the date the article was written)

See Also:
x86 Assembly

© 1999-2011 All rights reserved. Terms of Use Privacy Policy
Comments? Questions? Feedback? Click here!