反汇编解析一段小代码(面试题)

#include<iostream>
using namespace std;

struct S
{
	int a;
	int *p;
};
S s;
int main()
{
	int *temp=&s.a;
	temp[0]=10;
	temp[1]=12;
	s.p=temp;
	s.p[1]=20;
	s.p[0]=120;//报错
			
	return 0;
}

请问最后s.p[0]=120;为什么会报错？

原因如下:

temp=00477738
&temp=0012FF7C
s.p=00477738
&s.p=0047773C
s.p[0]=10//因为temp[0]=10
&s.p[0]=00477738
s.p[1]=4683576
&s.p[1]=0047773C

可见temp=&s.p[0]=s.p；&s.p=&s.p[1]

反汇编如下：

12:       int *temp=&s.a;
00401048   mov         dword ptr [ebp-4],offset s (00432e08)
13:       temp[0]=10;
0040104F   mov         eax,dword ptr [ebp-4]
00401052   mov         dword ptr [eax],0Ah
14:       temp[1]=12;
00401058   mov         ecx,dword ptr [ebp-4]
0040105B   mov         dword ptr [ecx+4],0Ch
15:       s.p=temp;
00401062   mov         edx,dword ptr [ebp-4]
00401065   mov         dword ptr [s+4 (00432e0c)],edx
16:       s.p[1]=20;
0040106B   mov         eax,[s+4 (00432e0c)]
00401070   mov         dword ptr [eax+4],14h
17:       s.p[0]=120;
00401077   mov         ecx,dword ptr [s+4 (00432e0c)]
0040107D   mov         dword ptr [ecx],78h
18:
19:       return 0;
00401083   xor         eax,eax
20:   }

00432E08  0A 00 00 00 14 00 00  ----------------------08 2E 43 变成了14 00 00(注意变量在内存中的存放方式，低位在前,高位在后)
00432E0F  00 00 00 00 00 00 00  
00432E16  00 00 24 E0 42 00 20

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