HOW TO OPTIMIZE FOR THE PENTIUM PROCESSOR ***************************************** Copyright (c) 1996 by Agner Fog Contents: 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 1. INTRODUCTION =============== 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 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. 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! 2. LITERATURE ============= A lot of useful literature can be downloaded for free from Intel's www site: www.intel.com You can find the documents you need by using the search facilities at: http://www-cs.intel.com/search.htm and: http://www-techdoc.intel.com/cgi-bin/taos.pl 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: http://www-techdoc.intel.com/cgi-bin/synopsis_spangle.pl?\docs\microproc\general\app_note\242816+0 A fancy tutorial named "Optimizing Applications for the Pentium Processor" is awailable at: http://www.intel.com/ial/processr/cbt.htm 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: http://www.intel.com/pc-supp/multimed/MMX 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 http://www.x86.org. 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 3. DEBUGGING ============ 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. 5. ALIGNMENT ============ 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. 6. CACHE ======== 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] 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. 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 POP register LEA, NOP 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: 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: 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 Examples: 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 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:, 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 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. 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 registers. Pairable instructions which read from memory, does some calculation, and stores the result in a register or flags, take 2 clock cycles. (read/modify instructions). 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 ----------------------|----------------------------------------------- Examples: 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 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 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. 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 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. 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 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 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 E: 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: 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 AND ECX,EAX XOR EAX,-1 AND EAX,EBX OR EAX,ECX 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. 13. PREFIXES ============ 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 effect. 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 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 conditional. Likewise, data adresses take less space if they can be expressed as a pointer and a displacement between -128 and +127. Example: MOV EBX,DS: / ADD EBX,DS: ; 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 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 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 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: 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. 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 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 ; 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. 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 ; clock cycle 4-9 Split up into: FILD [a] ; clock cycle 1-3 FILD ; clock cycle 2-4 FMUL ; clock cycle 5-7 In this example, you save two clocks by overlapping the two FILD instructions.