请大家看看，谢谢！！

[lys0796]

#include "stdafx.h"
#include "stdio.h"
#include "string.h"

int main(int argc, char* argv[])
{
unsigned char *p1;
unsigned long *p2;
p1=(unsigned char *)0x10000;
p2=(unsigned long *)0x10000;

printf("p1 %lx\n",p1);
printf("p2 %lx \n",p2);


printf("p1+5 %lx\n",p1+5);
printf("p2+5 %lx \n",p2+5);
printf("unsigned long %lx \n",5*sizeof(unsigned long));
printf("指针算个屁!\n");
return 0;

}


输出结果：

p1 10000
p2 10000
p1+5 10005
p2+5 10014///为什么？？
unsigned long 14///为什么？？
指针算个屁!

[kakaying]

p1 + 5 = 0x10000 + (sizeof(unsigned char) * 5) = 0x10000 + 0x5 = 0x10005
p2 + 5 = 0x10000 + (sizeof(unsigned long) * 5) = 0x10000 + 20(0x14) = 0x10014
5*sizeof(unsigned long)) = 5 * 4 = 20 = 0x14

[ForestDB]

指针的加减法是根据指针的类型(指针所指变量类型)来决定的.

[wbyhzh1111]

飘过

[bo.cui]

p2=(unsigned long *)0x10000; //强制指向地址空间[0x10000]

p2 + 5 = 0x10000 + (sizeof(unsigned long) * 5) = 0x10000 + 20(0x14) = 0x10014
//p2内为地址值10000，p2类型为unsigned long,加5表示向后移动5个unsigned long类型长度，既0x10014（具体值与unsigned long的长度有关，unsigned long长度多由系统决定）

5*sizeof(unsigned long)) = 5 * 4 = 20 = 0x14//同理

[bo.cui]

p2=(unsigned long *)0x10000; //强制指向地址空间[0x10000]

p2 + 5 = 0x10000 + (sizeof(unsigned long) * 5) = 0x10000 + 20(0x14) = 0x10014
//p2内为地址值10000，p2类型为unsigned long,加5表示向后移动5个unsigned long类型长度，既0x10014（具体值与unsigned long的长度有关，unsigned long长度多由系统决定）

5*sizeof(unsigned long)) = 5 * 4 = 20 = 0x14//同理

[bo.cui]

。。。。。。。
网速其慢，点一下居然发2次。 5555555555

[blh]

p1 10000
p2 10000
p1+5 10005
p2+5 10014///为什么？？ 因为 p2+5 == &p2[5],因为你定义的类型是long，每个long长度是4 bytes,所以p2+5偏移20个bytes,就是16进制的14
unsigned long 14///为什么？？ 20 就是16进制的14
指针算个屁!

建议学习一下
1、16进制、8进制、2进制和10进制数之间的转换
2、不同类型指针+/-操作的地址移动长度