memset movq给出段错误 [英] memset movq giving segfault

问题描述

我在movq(％rsi，％rcx)行出现段错误.

I am getting a segfault at the movq (%rsi, %rcx) line.

我知道您不能执行mem-> mem mov，所以我通过一个临时寄存器来完成它. (％rsi)，％r​​cx，然后在循环％rcx，(％rdi)中.这是我的代码:

I know you can't do mem->mem mov, so I did it through a temporary register. (%rsi), %rcx, then in the loop %rcx, (%rdi). Here is my code:

experimentMemset:   #memset(void *ptr, int value, size_t num)

                                 #%rdi     #%rsi        #%rdx


movq %rdi, %rax             #sets rax to the first pointer, to return later


.loop:
    cmp $0, (%rdx)          #see if num has reached 0
    je .end
    cmpb $0, (%rdi)         #see if string has ended also
    je .end

    movq %rsi, %rdi       #copies value into rdi

    inc %rdi        #increments pointer to traverse string
    dec %rdx        #decrements the count, aka num
    jmp .loop



.end:
     ret

推荐答案

如您所见，RDX拥有一个大小(一个整数)，而不是一个指针.它是通过值而不是通过引用传递的.

As you discovered, RDX holds a size (an integer count), not a pointer. It's passed by value, not by reference.

cmp $0, (%rdx)

不是比较寄存器，而是比较它所指向的位置.看来％rdx用作计数器，所以您应该比较寄存器本身.

compares not the register, but the location pointed by it. It seems that %rdx is used as a counter, so you should compare the register itself.

test %rdx,%rdx; je count_was_zero

还有其他错误，例如检查只写目标的内容是否为零，以及不将%sil存储到(%rdi)中.但这是在当前版本的问题中出现段错误的原因.

There are other bugs, like checking the contents of the write-only destination for zeros, and not storing %sil into (%rdi). But this was the cause of the segfault in the current version of the question.

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