如何读取objdump的汇编代码输出

时间:2018-10-24 16:15:38

标签: c disassembly objdump

我有一个C代码,可以交换两个数字。

#include<stdio.h>
void swap(int,int);        
void main( )
{
    int n1,n2;
    printf("Enter the two numbers to be swapped\n");
    scanf("%d%d",&n1,&n2);
    printf("\nThe values of n1 and n2 in the main function before calling the swap function are n1=%d n2=%d",n1,n2);
    swap(n1,n2);                                          
    printf("\nThe values of n1 and n2 in the main function after calling the swap function are n1=%d n2=%d",n1,n2);}

void swap(int n1,int n2)                           
{ 
    int temp;
    temp=n1;
    n1=n2;
    n2=temp;
    printf("\nThe values of n1 and n2 in the swap function after swapping are n1=%d n2=%d",n1,n2);
}

我已使用objdump对其进行了拆解,并试图找出交换操作在计算机级别如何发生。我认为这是交换功能。

000006b4 <swap>:
 6b4:   55                      push   %ebp
 6b5:   89 e5                   mov    %esp,%ebp
 6b7:   53                      push   %ebx
 6b8:   83 ec 14                sub    $0x14,%esp
 6bb:   e8 37 00 00 00          call   6f7 <__x86.get_pc_thunk.ax>
 6c0:   05 0c 19 00 00          add    $0x190c,%eax
 6c5:   8b 55 08                mov    0x8(%ebp),%edx
 6c8:   89 55 f4                mov    %edx,-0xc(%ebp)
 6cb:   8b 55 0c                mov    0xc(%ebp),%edx
 6ce:   89 55 08                mov    %edx,0x8(%ebp)
 6d1:   8b 55 f4                mov    -0xc(%ebp),%edx
 6d4:   89 55 0c                mov    %edx,0xc(%ebp)
 6d7:   83 ec 04                sub    $0x4,%esp
 6da:   ff 75 0c                pushl  0xc(%ebp)
 6dd:   ff 75 08                pushl  0x8(%ebp)
 6e0:   8d 90 c0 e8 ff ff       lea    -0x1740(%eax),%edx
 6e6:   52                      push   %edx
 6e7:   89 c3                   mov    %eax,%ebx
 6e9:   e8 72 fd ff ff          call   460 <printf@plt>
 6ee:   83 c4 10                add    $0x10,%esp
 6f1:   90                      nop
 6f2:   8b 5d fc                mov    -0x4(%ebp),%ebx
 6f5:   c9                      leave  
 6f6:   c3                      ret   

我想知道寄存器内的交换操作是如何发生的,我知道它必须是这样的。

push eax
mov eax, ebx
pop ebx

但是我看不到任何类似的东西。由于我是新手,所以有人可以帮助我如何了解情况如何。 Full output of the objdump is here.

1 个答案:

答案 0 :(得分:1)

要开始使用汇编语言,您可以检查以下链接:

http://patshaughnessy.net/2016/11/26/learning-to-read-x86-assembly-language