啦啦啦
本帖最后由 aynudx 于 2016-3-18 11:03 编辑
p1+5=0x801000+5*sizeof(char);
p2+5=0x810000+5*sizeof(long);
0x8010050x81000a
签到签到
【3.18签到】领金币
p1+5=0x801005
p2+5=0x810014
签到签到
签到签到先!
签到签到
unsigned char *p1;
unsigned long *p2;
p1=(unsigned char *)0x801000;
p2=(unsigned long *)0x810000;
请问p1+5=0x801005;
p2+5= 0x810014;