1.1 传送指令

1. push %ebp
2. mov %esp,%ebp
3. and $0xfffffff0,%esp //边界对齐指令，16字节对齐
4. sub $0x20,%esp
5. movl $0xa,0x10(%esp) //立即数10赋值给偏移量为0x10的基址寄存器esp
6. movl $0x9,0x14(%esp) //立即数9赋值
7. mov 0x10(%esp),%eax 
8. mov %eax,0x18(%esp) //立即数10赋值给0x18(%esp)
9. mov 0x14(%esp),%eax
10. mov %eax,0x1c(%esp) //立即数9赋值给0x1c(%esp)
11. mov 0x1c(%esp),%eax
12. mov %eax,0x8(%esp) //立即数9赋值给0x8(%esp)
13. mov 0x18(%esp),%eax
14. mov %eax,0x4(%esp) //立即数10赋值给0x4(%esp)
15. movl $0x0,(%esp)
16. call 41 <main+0x41>
17. leave
18. ret

参考代码：

#include"stdio.h"

int main(){

    int a,b,c,d;

    a=10,b=9;

    printf("评测结果：成功\n评测脚本：C\n返回结果：");//此行不在汇编代码中

    //**********Begin**********

    c=a;

    d=b;

    //**********End**********

    printf("%d,%d\n",c,d);

}

1.2 传送指令

1. mov %esp,%ebp
2. and $0xfffffff0,%esp
3. sub $0x20,%esp
4. movl $0xa,0x18(%esp) //立即数10赋值给偏移量为0x18的基址寄存器esp
5. mov 0x18(%esp),%eax
6. mov %eax,0x1c(%esp)
7. mov 0x1c(%esp),%eax
8. mov %eax,0x8(%esp) //立即数10赋值给0x8(%esp)
9. mov 0x18(%esp),%eax
10. mov %eax,0x4(%esp)
11. movl $0x0,(%esp)
12. call 31 <main+0x31>
13. leave
14. ret

参考代码： 

#include"stdio.h"

int main(){

    int a=10;

    int b;

    printf("评测结果：成功\n评测脚本：C\n返回结果：");//此行不在汇编代码中

    //**********Begin********

    b=a;

    printf("%d,%d\n",a,b);

    //**********End**********

}

2.1 加减运算指令

1. push %ebp
2. mov %esp,%ebp
3. and $0xfffffff0,%esp
4. sub $0x20,%esp movl
5. movl $0x6,0x10(%esp) //立即数6赋值给偏移量为0x10的基址寄存器esp
6. movl $0x64,0x14(%esp) //16进制 立即数为100
7. mov 0x14(%esp),%eax 
8. mov 0x10(%esp),%edx
9. sub %eax,%edx //将edx-eax的值赋值给edx
10. mov %edx,%eax
11. mov %eax,0x18(%esp) //将edx的值赋值给c
12. mov 0x14(%esp),%eax
13. mov 0x10(%esp),%edx
14. add %edx,%eax //将eax+edx的值赋值给edx
15. mov %eax,0x1c(%esp)
16. mov 0x1c(%esp),%eax
17. mov %eax,0x8(%esp) //将eax的值赋值给c
18. mov 0x18(%esp),%eax
19. mov %eax,0x4(%esp)
20. movl $0x0,(%esp)
21. call 4f <main+0x4f>
22. leave
23. ret

参考代码 ：

#include"stdio.h"

int main(){

    int a,b,c,d;

    a=6,b=100;

    printf("评测结果：成功\n评测脚本：C\n返回结果：");//此行不在汇编代码中

    //**********Begin**********

    c=a-b;

    d=a+b;

    //**********End**********

    printf("%d,%d\n",c,d);

}

3.1 整数的乘法指令

1. push %ebp
2. mov %esp,%ebp
3. and $0xfffffff0,%esp
4. sub $0x20,%esp
5. movl $0x3,0x10(%esp) //立即数3赋值给偏移量为0x10的基址寄存器esp
6. mov 0x10(%esp),%eax
7. sub $0x1,%eax //将eax的值-1赋给eax
8. imul 0x10(%esp),%eax //将eax与0x10(%esp)内的值相乘赋给eax
9. mov %eax,0x14(%esp) //赋值给b
10. mov 0x10(%esp),%eax 
11. imul 0x14(%esp),%eax //将c与0x10(%esp)内的值相乘赋给eax
12. mov %eax,0x18(%esp) //赋给c
13. mov 0x18(%esp),%eax
14. add $0x2,%eax //d+2的值赋给eax
15. imul 0x10(%esp),%eax //eax与0x10(%esp)内的值相乘赋给eax
16. mov %eax,0x1c(%esp) //赋给d
17. mov 0x1c(%esp),%eax
18. mov %eax,0x8(%esp)//赋给d
19. mov 0x18(%esp),%eax
20. mov %eax,0x4(%esp) 
21. movl $0x0,(%esp)
22. call 56 <main+0x56>
23. leave
24. ret

 参考代码：

#include"stdio.h"

int main(){

    int a,b,c,d;

    a=3;

    printf("评测结果：成功\n评测脚本：C\n返回结果：");//此行不在汇编代码中

    //**********Begin**********

    b=a*(a-1);

    c=a*b;

    d=a*(c+2);

    //**********End**********

    printf("%d,%d\n",c,d);

}

4.1 控制转移指令

1. 00000000 <sum>:
2. 0: 55              push %ebp
3. 1: 89 e5            mov %esp,%ebp
4. 3: 83 ec 10          sub $0x10,%esp //分配十字节空间
5. 6: c7 45 fc 00 00 00 00  movl $0x0,-0x4(%ebp) //s =0
6. d: c7 45 f8 00 00 00 00  movl $0x0,-0x8(%ebp) //i =0
7. 14: eb 18           jmp 2e <sum+0x2e>
8. 16: 8b 45 f8         mov -0x8(%ebp),%eax //
9. 19: 8d 14 85 00 00 00 00 lea 0x0(,%eax,4),%edx //a[i]的值赋给edx
10. 20: 8b 45 08         mov 0x8(%ebp),%eax //
11. 23: 01 d0           add %edx,%eax
12. 25: 8b 00           mov (%eax),%eax
13. 27: 01 45 fc         add %eax,-0x4(%ebp) //s的值+a[i]赋给s
14. 2a: 83 45 f8 01       addl $0x1,-0x8(%ebp) //i的值+1赋给i
15. 2e: 8b 45 f8         mov -0x8(%ebp),%eax //i的值赋给eax
16. 31: 3b 45 0c         cmp 0xc(%ebp),%eax //比较i跟n的值；
17. 34: 7c e0           jl 16 <sum+0x16>
18. 36: 8b 45 fc         mov -0x4(%ebp),%eax
19. 39: c9             leave
20. 3a: c3 ret

 参考代码：

#include"stdio.h"

int sum(int a[],int n){

    //**********Begin**********

    int s = 0;

    for (int i = 0;i<n; i++) {

        s += a[i]; 

    }

   return s;

    //**********End**********

}