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