反汇编一个简单的C程序并分析

#反汇编一个简单的C程序并分析

C 源码:

int g(int x)
{
    return x+1;
}

int f(int x)
{
    return g(x);
}

int main(void)
{
    return f(2) + 3;
}

汇编源码:

1 g:
2    pushl    %ebp
3    movl    %esp, %ebp
4    movl    8(%ebp), %eax
5    addl    $1, %eax
6    popl    %ebp
7    ret
8 f:
9    pushl    %ebp
10    movl    %esp, %ebp
11    subl    $4, %esp
12    movl    8(%ebp), %eax
13    movl    %eax, (%esp)
14    call    g
15    leave
16    ret
17 main:
18    pushl    %ebp
19    movl    %esp, %ebp
20    subl    $4, %esp
21    movl    $2, (%esp)
22    call    f
23    addl    $3, %eax
24    leave
25    ret

执行过程:(从 main 开始)

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July 03,2015