栽了两次的题--不要忘记

#include<iostream>
using namespace std;
struct A
{
 char c;
 short s;
 char cc;
 long l;
}*pAbc;
int main()
{
pAbc=(struct A *)0x100000;
printf("%x\n",pAbc+0x100);
printf("%x\n",(unsigned long)pAbc+0x100);
printf("%x\n",(char *)pAbc+0x100);
printf("%x\n",(unsigned long *)pAbc+0x100);
}
 pAbc+0x100 = 0x100000 + 0x100 * 12 = 0x100C00;
  (ULONG)pAbc + 0x100 = 0x100000 + 0x100 = 0x100100;
  (ULONG *)pAbc + 0x100 = 0x100000 + 0x100 * 4 = 0x100400;
  (char *)pAbc + 0x100 = 0x100000 + 0x100 * 1 = 0x100100;

sizeof(struct A)=12, pAbc是一个指向这种类型结构体的一个指针,pAbc+0x100就相当于指针pAbc偏移0x100个单元的位置, 每个单元式12,所以就乘以12。

(ULONG *)pAbc + 0x100 = 0x100000 + 0x100 * 4 = 0x100400;
这里pAbc已经强制转型为(ULONG *), 这时单元的大小发生了变化。变成sizeof(ULONG)=4了,所以乘以4.
*1的同理。

posted on 2011-09-15 15:20  原来...  阅读(2105)  评论(0)    收藏  举报

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