一段程序，给我解释下吧

jiangmingfu010 2008-10-18 12:04:13

unsigned char *p1;
unsigned long *p2;
p1=(unsigned char *)0x801000; //此处的转化到底是什么意思啊
p2=(unsigned long *)0x810000;

printf("%x\n",p1+5);
printf("%x\n",p2+5);

为什么输出的是801005
810014

...全文

86 4 打赏收藏转发到动态举报

写回复

用AI写文章

4 条回复

切换为时间正序

请发表友善的回复…

发表回复

jiangmingfu010 2009-09-22

打赏
举报

unsigned char *p1;
unsigned long *p2;
p1=(unsigned char *)0x801000; //此处的转化到底是什么意思啊. 就是令p1指针指向0X801000这个地址.
p2=(unsigned long *)0x810000;

printf("%x\n",p1+5);//0X801000+5*sizeof(unsigned char)
printf("%x\n",p2+5);//0X810000+5*sizeof(unsigned long)

phz1985 2008-10-18

打赏
举报

结帖这么快.LZ是个好人.^_^

jiangmingfu010 2008-10-18

打赏
举报

[Quote=引用 1 楼 phz1985 的回复:]
C/C++ code
unsigned char *p1;
unsigned long *p2;
p1=(unsigned char *)0x801000; //此处的转化到底是什么意思啊. 就是令p1指针指向0X801000这个地址.
p2=(unsigned long *)0x810000;

printf("%x\n",p1+5);//0X801000+5*sizeof(unsigned char)
printf("%x\n",p2+5);//0X810000+5*sizeof(unsigned long)
[/Quote]
谢谢了，分全给你了

phz1985 2008-10-18

打赏
举报



  unsigned char *p1;

  unsigned long *p2;

  p1=(unsigned char *)0x801000;  //此处的转化到底是什么意思啊. 就是令p1指针指向0X801000这个地址.

  p2=(unsigned long *)0x810000;



  printf("%x\n",p1+5);//0X801000+5*sizeof(unsigned char)

  printf("%x\n",p2+5);//0X810000+5*sizeof(unsigned long)