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| python3 对 64 位的二进制进行异或的运算后，结果变成了 32 位的，这是什么逻辑？ Python • jiangjunhui4000 • Sep 21, 2018 • Lastly replied by ysc3839 | 4 |
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Sep 21, 2018 Replied to a topic by jiangjunhui4000 › Python › python3 对 64 位的二进制进行异或的运算后，结果变成了 32 位的，这是什么逻辑？ |
我指的是二进制的位数，如一下两个 64 位二进制
hash1 = '0b0010000110100001101101010101111110001001100001101011111110101000'
hash2 = '0b0010000110101001101101010101111110101001100001101011101110101000'
In [51]: bin(int(hash1,2)^int(hash2,2))
Out[51]: '0b1000000000000000000000100000000000000000010000000000' ##52 位
异或之后的位数不是 64 位了，不清楚中间的换算逻辑是怎样的
hash1 = '0b0010000110100001101101010101111110001001100001101011111110101000'
hash2 = '0b0010000110101001101101010101111110101001100001101011101110101000'
In [51]: bin(int(hash1,2)^int(hash2,2))
Out[51]: '0b1000000000000000000000100000000000000000010000000000' ##52 位
异或之后的位数不是 64 位了，不清楚中间的换算逻辑是怎样的
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