摘要: #include <stdio.h>int main(){ int n = 0x22445566; int b = ((n & 0x000000ff) << 24)|((n & 0x0000ff00) << 8) | ((n & 0x00ff0000) >> 8) | ((n &0xff000000) >> 24); printf("n = %X\nb = %X\n", n, b);} 阅读全文
posted @ 2013-01-23 12:51 Arya_yu 阅读(690) 评论(0) 推荐(0) 编辑