简单c程序汇编代码分析

c源代码如下：

#include <stdio.h>
#include <stdlib.h>

int test()
{
	int arg1 = 3;
	int arg2 = 7;

	int diff = arg2 - arg1;

	return diff;	
}

int main()
{
	int res;

	res = test();

	printf("res=%d\n", res);	

	return 0;
}

首先使用gcc test.c -g对上面的代码进行编译，自动生成目标文件a.out；然后使用objdump -S a.out查看汇编代码，截取片断如下：

int test()
{
 80483e4:	55                   	push   %ebp		保存帧指针，帧指针入栈
 80483e5:	89 e5                	mov    %esp,%ebp	使帧指针指向栈指针
 80483e7:	83 ec 10             	sub    $0x10,%esp	申请16字节的栈空间
	int arg1 = 3;
 80483ea:	c7 45 fc 03 00 00 00 	movl   $0x3,-0x4(%ebp)	把3入栈
	int arg2 = 7;
 80483f1:	c7 45 f8 07 00 00 00 	movl   $0x7,-0x8(%ebp)	把7入栈

	int diff = arg2 - arg1;
 80483f8:	8b 45 fc             	mov    -0x4(%ebp),%eax	取出3，放入寄存器%eax
 80483fb:	8b 55 f8             	mov    -0x8(%ebp),%edx	取出7，放入寄存器%edx
 80483fe:	89 d1                	mov    %edx,%ecx	把寄存器%edx的值交给寄存器%ecx
 8048400:	29 c1                	sub    %eax,%ecx	寄存器%ecx的值减去寄存器%eax的值，再放入寄存器%ecx
 8048402:	89 c8                	mov    %ecx,%eax	把寄存器%ecx的值交给寄存器%eax
 8048404:	89 45 f4             	mov    %eax,-0xc(%ebp)	寄存器%eax的值保存在栈上，diff是局部变量，他的值是存储在栈上的

	return diff;	
 8048407:	8b 45 f4             	mov    -0xc(%ebp),%eax	从栈取出diff，交给寄存器%eax，这样寄存器%eax保存的就是函数返回值了
}
 804840a:	c9                   	leave  			准备栈，这里应该是恢复到之前的栈，即要恢复%esp和%ebp
 804840b:	c3                   	ret    			返回到call调用之后的那个地址继续执行，即返回到804841a

0804840c <main>:


int main()
{
 804840c:	55                   	push   %ebp
 804840d:	89 e5                	mov    %esp,%ebp
 804840f:	83 e4 f0             	and    $0xfffffff0,%esp
 8048412:	83 ec 20             	sub    $0x20,%esp
	int res;

	res = caller();
 8048415:	e8 ca ff ff ff       	call   80483e4 <test>	调用test函数
 804841a:	89 44 24 1c          	mov    %eax,0x1c(%esp)	将返回值入栈

	printf("res=%d\n", res);	
 804841e:	b8 00 85 04 08       	mov    $0x8048500,%eax
 8048423:	8b 54 24 1c          	mov    0x1c(%esp),%edx
 8048427:	89 54 24 04          	mov    %edx,0x4(%esp)
 804842b:	89 04 24             	mov    %eax,(%esp)
 804842e:	e8 e9 fe ff ff       	call   804831c <printf@plt>

	return 0;
 8048433:	b8 00 00 00 00       	mov    $0x0,%eax
}
 8048438:	c9                   	leave  
 8048439:	c3                   	ret    
 804843a:	90                   	nop