20
在对齐为4的情况下
struct BBB
{
long num; 4
char *name; 4
short int data; 2
char ha; 1
short ba[5];
}*p;
p=0x1000000;
p+0x200=____;
(Ulong)p+0x200=____;
(char*)p+0x200=____;
希望各位达人给出答案和原因,谢谢拉
解答:假设在32位CPU上,
sizeof(long) = 4 bytes
sizeof(char *) = 4 bytes
sizeof(short int) = sizeof(short) = 2 bytes
sizeof(char) = 1 bytes
由于是4字节对齐,
sizeof(struct BBB) = sizeof(*p)
= 4 + 4 + 2 + 1 + 1/*补齐*/ + 2*5 + 2/*补齐*/ = 24 bytes (经Dev-C++验证)
p=0x1000000;
p+0x200=____;
= 0x1000000 + 0x200*24
这个懂了 每次p+1其实都是移动一个结构体大小的长度
=================================
(Ulong)p+0x200=____;
= 0x1000000 + 0x200
(char*)p+0x200=____;
= 0x1000000 + 0x200*4
这两个没看懂啊 为什么是这样了?????
求解释