在汇编 x86 中计算 LCM [英] Calculating LCM in assembly x86

问题描述

我有以下汇编代码

.global _start

.section .text
_start:
    movq a, %rax
    movq b, %rbx
    imul %rbx, %rax
    cmp %rbx, %rax
    je gcd_calculated
    ja l1
    sub %rax, %rbx
    jmp _start
l1:
    sub %rbx, %rax
    jmp _start
gcd_calculated:
    div %rax
    movq %rax, (c)
    

a,b 是我需要计算它们的 lcm 的四边形，我需要将结果分配给 c

a,b are quads that I need to calculate their lcm and I need to assign the result to c

我在上面的代码中得到错误的结果，我不知道为什么.

I get wrong results with the above code and I can't spot why.

通常，我在 lcm = (a*b)/gcd 上中继，所以我将 a*b 存储在 %rax 中，然后计算 gcd将存储在 %eax 并最终在它们之间进行划分.

generally, i'm relaying on the the lcm = (a*b)/gcd so I store a*b in %rax and then calculate the gcd which will be stored at %eax and finally divide between them.

推荐答案

就像 1201ProgramAlarm 说的，你在每次迭代中计算和比较 a 和 b 的原始值你的循环.如果您想遵循当前的方法，则必须在每次循环之前将这些值存储在内存中.

Like 1201ProgramAlarm said, you are calculating and comparing the original values of a and b in each iteration of your loop. If you want to follow your current approach, you would have to store these values in memory each time before looping.

我实际上要做的只是先进行 gcd 计算，然后在最后计算 (a*b)/gcd.

What I would actually do instead is just do the gcd calculation first and then calculate (a*b)/gcd at the end.

_start:
    movq a(%rip), %rbx
    movq b(%rip), %rcx
gcd:
    cmp %rbx, %rcx
    je gcd_calculated
    ja l1
    sub %rcx, %rbx
    jmp gcd
l1:
    sub %rbx, %rcx
    jmp gcd
gcd_calculated:
    movq a(%rip), %rax
    xor %rdx, %rdx
    div %rbx
    imul b(%rip), %rax
    movq %rax, c(%rip)

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