[code:c]
#include
int main()
{
int a[5]={1,2,3,4,5};
int *ptr1=(int *)(&a+1);
int *ptr2=(int *)((int)a+1);
printf("%x,%x",ptr1[-1],*ptr2);
return 0;
}
[/code]
37楼 kingqin 大牛给的答案不错,学习了。
[code:c]
#include
int main(void)
{
int a[5]={1,2,3,4,5};
//假设a的地址为0x80000000,sizeof(int)=4
//假设是little-endian,则0x80000000-0x80000013的数据为:
// addr: 00 01 02 03 04 05 06 07
// data: 01 00 00 00 02 00 00 00
// addr: 08 09 0a 0b 0c 0d 0e 0f 10 11 12 13
// data: 03 00 00 00 04 00 00 00 05 00 00 00
int *ptr1=(int *)(&a+1);
//ptr1 = 0x80000000+20=0x80000014
int *ptr2=(int *)((int)a+1);
//ptr2 = 0x80000000+1 = 0x80000001
//&ptr1[-1] = 0x80000014-4 = 0x80000010 ;
//so ptr1[-1] = *(int *)0x80000010 = 5
//*ptr2 = *(int *)0x80000001 = 0x02000000
printf("%x,%x",ptr1[-1],*ptr2);
return 0;
}
[/code]
csdn 上很多人说这个很简单,意思是其实就是 我太菜了! 怎么办啊 ?!!