02 2012 档案

Impossible Codification
摘要：在iq的blog上看到这样一段代码:int t[] = {0x4845A956, 0x586DEE32, 0x7E6B9933, 0x0D059D58, 0};int ch = t[0] + t[1] + t[2] + t[3];t[0]^=ch;t[1]^=ch;t[2]^=ch;t[3]^=ch;char *str = (char*)t;把str打印出来是这样的：Egad! It WORKS!!而如果将t的改为这样：int t[] = {0x151ba3a, 0x10abc1a, 0x118a113, 0x1e08bc0b, 0 };那么此时str为: Hi, how are you?.看 阅读全文
posted @ 2012-02-19 18:44 atyuwen 阅读(2029) 评论(1) 推荐(0) 编辑