这台机器代码在C中会怎么样？

Question

我有这个机器代码,我在过去4天一直在努力,但我似乎无法掌握它.

    pushl %ebp
    movl %esp, %ebp
    jmp .L3
L4:
    addl $3,8(%ebp)

L3:
    movl 8(%ebp),%eax // having issues here
    movzbl (%eax),%eax//here
    cmpb $1,%al// and here
    je .L5
    cmpl 12(%ebp),%eax
    jne .L4
    movl $8,%eax
.L5: leave
     ret

这是我对解决方案的尝试:

int main(int s, char *t){
    while (s != 1) {
        if (s == *t){
            s = 8;
        }
        else{s+=3;}
    }
    return;
}

有人能告诉我,我是否正确接近这个？如果没有帮助我指向正确的方向？

Answer 1

该函数看起来是__cdecl,因为它引用8(%ebp)和12(%ebp)不引用任何一个,这表明它们是参数.叫他们arg1和arg2分别.

我们可以注释程序集:

    /* Function prologue */
    pushl %ebp
    movl %esp, %ebp

    jmp .L3
L4:
    /* Move to next loop iteration */
    addl $3, arg1

L3:
    /* Copy value of arg1 to %eax */
    movl arg1, %eax
    /* (%eax) means to dereference %eax (as a memory address).
       movzbl means to load a single byte and zero-extend to the rest of the register.
       Therefore, the byte at address %eax is put into %eax. */
    movzbl (%eax), %eax
    /* Compare the low byte of %eax against 1 */
    cmpb $1, %al
    /* If equal, exit (%eax = 1) */
    je .L5
    /* Compare %eax against arg2 */
    cmpl arg2, %eax
    /* If not equal, keep looping */
    jne .L4

    /* Otherwise, if it was equal, we quit and return %eax = 8 */
    movl $8,%eax
.L5: leave
     ret

在C代码中,这变成了

int fn(unsigned char *ptr, int sentinel) {
    while(1) {
        unsigned char c = *ptr;
        if(c == 1) return c;
        if(c == sentinel) return 8;
        ptr += 3;
    }
}