闲着蛋疼看下a++的过程

赋值过程

int a = 1;
int b = a++;

x86 反汇编:

  int a = 1;
00D06428 C7 45 F8 01 00 00 00 mov         dword ptr [a],1  
  int b = a++;
00D0642F 8B 45 F8             mov         eax,dword ptr [a]  
00D06432 89 45 EC             mov         dword ptr [b],eax  先赋值
00D06435 8B 4D F8             mov         ecx,dword ptr [a]  
00D06438 83 C1 01             add         ecx,1  
00D0643B 89 4D F8             mov         dword ptr [a],ecx  在修改

传递过程

int a = 1;
printf("%d\n", a++);

x86反汇编:

  int a = 1;
003E6428 C7 45 F8 01 00 00 00 mov         dword ptr [a],1  
  printf("%d\n", a++);
003E642F 8B 45 F8             mov         eax,dword ptr [a]  
003E6432 89 85 30 FF FF FF    mov         dword ptr [ebp-0D0h],eax  存了旧值
003E6438 8B 4D F8             mov         ecx,dword ptr [a]  
003E643B 83 C1 01             add         ecx,1  
003E643E 89 4D F8             mov         dword ptr [a],ecx  a已被改变
003E6441 8B 95 30 FF FF FF    mov         edx,dword ptr [ebp-0D0h]  取了旧值
003E6447 52                   push        edx  
003E6448 68 30 8C 44 00       push        offset string "%d\n" (0448C30h)  
003E644D E8 DE 9D FF FF       call        _printf (03E0230h)  调用函数
003E6452 83 C4 08             add         esp,8  

保存旧值，修改值，传递旧值