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与(&)或(|)非(~)实现异或(^)

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Bruce ACG 最后由编辑
一道很基础但有点意思的算法题，我的思路是
```
int my_xor(int a, int b)
{
        return (a&~b)|(~a&b);
}
```
不过网上还有一种比较奇妙的解法，返回(a|b)&(~a|~b)
结果确实正确，不过很好奇他怎么想到的...
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