学步园

写出下面程序的输入结果：

#include <stdio.h>
struct BBB
{
	long num;
	char *name;
	short int data;
	char ha;
	short ba[5];
}*p;

int main()
{
	p = 0x1000000;
	printf(" p+0x200=0x%08x\n", p+0x200);
	printf(" (Ulong)p+0x200=0x%08x\n", (unsigned long)p+0x200);
	printf("(char*)p+0x200=0x%08x\n", (char*)p+0x200);
	printf("(short*)p+0x200=0x%08x\n", (unsigned short*)p+0x200);
	printf("(int*)p+0x200=0x%08x\n", (unsigned int*)p+0x200);

	return 0;
}

运行结果：
p+0x200=0x01003000
 (Ulong)p+0x200=0x01000200
(char*)p+0x200=0x01000200
(short*)p+0x200=0x01000400
(int*)p+0x200=0x01000800

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

(char*)p+0x200 = 0x1000000 + 0x200 * 1 = 0x01000200
(short*)p+0x200 = 0x1000000 + 0x200 * 2  = 0x01000400
(int*)p+0x200 = 0x1000000 + 0x200 * 4 = 0x01000800

结论：

某一类型的指针 +  一个常数 =  这个常数 * 指针所指区域的大小。