看到这样一个问题：只用位运算实现比较两整数大小
自己尝试了一下，写了个强行加入位运算的，但是有四则运算:

int i = 0xffffffff;
return (!((unsigned)(a-b)>>(sizeof(int)*8-1))*i)&a|(!((unsigned)(b-a)>>(sizeof(int)*8-1))*i)&b;

直接pass，;(

发现知乎上有这个解答：
https://www.zhihu.com/question/44356016

仰望高端玩家233333333，还是要吸收消化一下。

以下代码属于转载，特此申明，；）

作者：夏洋
链接：
https://www.zhihu.com/question/44356016/answer/97251524
来源：知乎
著作权归作者所有，转载请联系作者获得授权。

int compare(uint32_t a, uint32_t b) {
  uint32_t diff = a ^ b;
  if (!diff) return 0;

    // 001xxxxx -> 00100000
    diff |= diff >> 1;
    diff |= diff >> 2;
    diff |= diff >> 4;
    diff |= diff >> 8;
    diff |= diff >> 16;
    diff ^= diff >> 1;
    return a & diff ? 1 : -1;
}

理解：
1.a==b 异或则为0
2.a < b 从高位向低位数起，二者第一次出现不同的二进制位时，a在该位处的值一定为0，b在该位处的值一定为1,反之亦然。

diff = a ^ b ，找diff中为1的那一位，a | 0 不变，a | 1 置1 ，那么>>16之后，为1的那一位还是1，为1之后的所有位都被置1

    diff |= diff >> 1;
    diff |= diff >> 2;
    diff |= diff >> 4;
    diff |= diff >> 8;
    diff |= diff >> 16;

为1的那一位置1，其他所有位因为异或，置0

    diff ^= diff >> 1;

检验a在该位是0还是1，是0的话说明a小，是1的话说明a大

    a & diff

以上。