instructions

Why are there no NAND, NOR and XNOR instructions in X86?

﹥>﹥吖頭↗ 提交于 2021-02-10 17:48:01
问题 They're one of the simplest "instructions" you could perform on a computer (they're the first ones I'd personally implement) Performing NOT(AND(x, y)) doubles execution time AND dependency chain length AND code size BMI1 introduced "andnot" which is a meaningful addition that is a unique operation - why not the ones in the title of this question? You usually read answers among the lines of "they take up valuable op-code space" but then I look at all of the kmask operations introduced with

Why are there no NAND, NOR and XNOR instructions in X86?

假如想象 提交于 2021-02-10 17:46:36
问题 They're one of the simplest "instructions" you could perform on a computer (they're the first ones I'd personally implement) Performing NOT(AND(x, y)) doubles execution time AND dependency chain length AND code size BMI1 introduced "andnot" which is a meaningful addition that is a unique operation - why not the ones in the title of this question? You usually read answers among the lines of "they take up valuable op-code space" but then I look at all of the kmask operations introduced with

Profiling instructions

独自空忆成欢 提交于 2021-02-08 06:26:13
问题 I want to count several cpu instructions in my code. e.g. I would like to know how many additions, how many multiplications, how many float operations, how many branches my code executes. I currently use gprof under Linux for profiling my c++ code but it only gives the number of calls to my functions, and I manually estimate the number of instructions. Are there any tools that might do the trick for me? Maybe some virtual machine? 回答1: You may be able to use Valgrind's Callgrind with the -

ARM assembly cannot use immediate values and ADDS/ADCS together

坚强是说给别人听的谎言 提交于 2021-02-07 12:52:26
问题 I am currently trying to speed up some of my C functions on a Cortex-M0 (Freescale KL25Z) using assembly. I get a problem with this minimal test program: @.syntax unified .cpu cortex-m0 .text .global test .code 16 test: mov r0, #0 adds r0, r0, #1 bx lr When I try to assemble my .s file to a .o file, I get this error $ arm-none-eabi-as test.s -o test.o test.s: Assembler messages: test.s:8: Error: instruction not supported in Thumb16 mode -- `adds r0,r0,#1' The error message doesn't make sense

Difference between memory and register

不羁的心 提交于 2021-02-07 04:35:13
问题 I saw assembly code like, MOV [EAX], EBX the above line, They are mentioned [EAX] is memory and EBX is Register. So, here what is the difference between [EAX] and EBX . What will happen in above instruction. 回答1: In this syntax, brackets around a register means a memory location is used (as source or destination, according to the instruction) with starting address specified at the register (EAX in your case). For example, if EAX contained 1344 before the instruction, value from EBX is copied

What's the difference between `add al, 0` and `add al, '0'`?

孤街浪徒 提交于 2021-02-05 10:40:39
问题 So I've been reading some assembly code for learning purposes and have come across these two instructions: add register, value add register, 'value' ; Where the value is now in single quotes What's the difference between the two? Before I get flamed if this happens to be a duplicate. I've asked here as I don't know exactly what to Google to answer this question. 回答1: Depending upon the assembler, enclosing a character in single quotes may request that the compiler use the ASCII code of that

Instructions to copy the low byte from an int to a char: Simpler to just do a byte load?

≯℡__Kan透↙ 提交于 2021-02-05 06:39:28
问题 I was reading a text book and it has an exercise that write x86-64 assembly code based on C code //Assume that the values of sp and dp are stored in registers %rdi and %rsi int *sp; char *dp; *dp = (char) *sp; and the answer is: //first approach movl (%rdi), %eax //Read 4 bytes movb %al, (%rsi) //Store low-order byte I can understand it but just wondering can't we do sth simple in the first place as: //second approach movb (%rdi), %al //Read one bytes only rather than read all four bytes movb

Instruction FYL2XP1

自闭症网瘾萝莉.ら 提交于 2021-01-23 06:33:47
问题 I'm wondering why the instruction FYL2XP1 on x86-architecture computes exactly the mathematical formula y · log 2 ( x + 1). What's special with this formula? 回答1: The y operand is usually a compile time constant, for the moment forget about the x + 1 . Since log_b(x) = log_b(2) * log_2(x) the instruction allows to compute the logarithm in any base of x + 1 . Note that log_b(2) is a constant since it is seldom necessary to compute the logarithm with a degree of freedom in the base. FYL2XP1 and

Difference between an instruction and a micro-op

一曲冷凌霜 提交于 2021-01-20 19:30:51
问题 What is the difference between a machine instruction and a micro-op? I found a following definition here: A small, basic instruction, used in series to make up a high-level machine instruction Here is what I found on Wikipedia In computer central processing units, micro-operations (also known as a micro-ops or μops) are detailed low-level instructions used in some designs to implement complex machine instructions (sometimes termed macro-instructions in this context) Am I correct in

Observing x86 register dependencies

。_饼干妹妹 提交于 2021-01-05 07:16:24
问题 Are there any other processor registers (e.g. flags) besides the architectural registers (eax, ebx,.) in x86 for which RAW dependencies need to be enforced by the scoreboard in pipelined processors? 回答1: Literally every register guarantees that if you write it, later instructions will read the new value. x86 is defined in terms of serial execution; pipelining and out-of-order exec need to preserve that illusion for everything , including segment registers, FP rounding modes, control and debug