汇编学习一

随便写了个代码分析汇编

源代码

#include<stdio.h>
int sum(int a,int b)
{
    int tmp=0;
    tmp=a+b;
    return tmp;
}
 
int main()
{
    int a=10;
    int b=20;
    int ret=0;
 
    ret=sum(a,b);
    printf("ret=%d\n",ret);
    return 0;

}

代码比较简单，但分析起来却对我这个小白极不友善，话不多说，直接上汇编

　　(1)main()函数汇编代码

 1     push   rbp                                                 # rbp入栈
 2     mov    rbp,rsp                                             # rsp的值放入rbp
 3     sub    rsp,0x30                                            # rsp-=0x30
 4     call   0x402130 <__main>                                   # 调用main()
 5     mov    DWORD PTR [rbp-0x4],0xa                             # [rbp-0x4]地址(a)
 6     mov    DWORD PTR [rbp-0x8],0x14                            # [rbp-0x8]地址(b)
 7     mov    DWORD PTR [rbp-0xc],0x0                             # [rbp-0xc]地址ret
 8     mov    edx,DWORD PTR [rbp-0x8]                             # rbp-0x8为地址里的值赋予edx,即edx=20
 9     mov    eax,DWORD PTR [rbp-0x4]                             # eax=10
10     mov    ecx,eax                                             # ecx=10
11     call   0x401530 <sum(int, int)>
12     mov    DWORD PTR [rbp-0xc],eax                             # eax是sum()返回值 eax=30
13     mov    eax,DWORD PTR [rbp-0xc]
14     mov    edx,eax                                             # edx=30
15     lea    rcx,[rip+0x2a69]                                    # 0x404000
16     call   0x402b48 <printf>
17     mov    eax,0x0                                             # eax=0
18     add    rsp,0x30                                            # 释放
19     ret

　　（2）sum()汇编代码

 1     push   rbp
 2     mov    rbp,rsp
 3     sub    rsp,0x10                                          # 生成栈空间（栈帧）
 4     mov    DWORD PTR [rbp+0x10],ecx                          # 已知ecx=10
 5     mov    DWORD PTR [rbp+0x18],edx                          # 已知edx=20
 6     mov    DWORD PTR [rbp-0x4],0x0                          
 7     mov    edx,DWORD PTR [rbp+0x10]                          # edx=10
 8     mov    eax,DWORD PTR [rbp+0x18]                          # eax=20
 9     add    eax,edx                                           # eax=10+20
10     mov    DWORD PTR [rbp-0x4],eax                           # 赋值                           
11     mov    eax,DWORD PTR [rbp-0x4]                           # 赋值
12     add    rsp,0x10                                          # rsp=rbp释放栈空间
13     pop    rbp                                               #
14     ret

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