在汇编语言分段错误 [英] Segmentation Fault in Assembly Language

问题描述

我learningn AT＆amp; T公司的x86汇编语言。我试图写一个汇编程序它接受一个整数n，然后返回结果（N / 2 + N / 3 + N / 4）。这里是我做了什么：

I am learningn AT&T x86 assembly language. I am trying to write an assembly program which takes an integer n, and then return the result (n/2+n/3+n/4). Here is what I have done:

.text
.global _start
_start:
    pushl $24
    call profit
    movl %eax, %ebx
    movl $1, %eax
    int $0x80

profit:
    popl %ebx
    popl %eax
    mov $0, %esi
    movl $4, %ebp
    div %ebp
    addl %eax, %esi
    movl %ecx, %eax
    movl $3, %ebp
    div %ebp
    addl %eax, %esi
    movl %ecx, %eax
    movl $2, %ebp
    div %ebp
    addl %eax, %esi
    movl %esi, %eax
    cmpl %ecx, %esi
    jg end
    pushl %ebx
    ret

end:
    mov %ecx, %eax
    ret

问题是我得到分段错误。问题出在哪里？

The problem is I am getting segmentation fault. Where is the problem?

推荐答案

我觉得code在这里失败：

I think the code fails here:

_start:
    pushl $24
    call profit
    movl %eax, %ebx
    movl $1, %eax
    int $0x80

profit:
    popl %ebx
    popl %eax

所以，你推$ 24个（4字节），然后呼叫利润，这推 EIP ，并跳转到利润。然后你弹出 EIP 的价值为 EBX 和值 $ 24个到 EAX 。

So, you push $24 (4 bytes) and then call profit, which pushes eip and jumps to profit. Then you pop the value of eip into ebx and the value $24 into eax.

然后，在最后，如果 JG结束分支结束：，那么栈将不会举行一个有效的返回地址， RET 将失败。你可能需要 pushl％EBX 有太多。

Then, in the end, if jg end branches to end:, then the stack won't hold a valid return address and ret will fail. You probably need pushl %ebx there too.

    cmpl %ecx, %esi
    jg end
    pushl %ebx
    ret

end:
    mov %ecx, %eax
    ; `pushl %ebx` is needed here!
    ret

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