Why AND(&) operator used calculating final index in Hashmap why not modulo(%) operator i.e hashValue & ()

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英文:

Why AND(&) operator used calculating final index in Hashmap why not modulo(%) operator i.e hashValue & ()

问题

我正在尝试理解 HashMap 的实现。我发现在计算键哈希值之后,最终的哈希值是使用按位与操作符(hashValue & (n-1))生成的,其中 n 是桶的大小。是否有人可以解释一下,为什么没有使用取模运算,取模运算也可以保证输出范围在桶的大小内呢?

英文:

I am trying to understand HashMap implementation. I found after calculating key hash value, final hash value is generated using AND operator (hashValue & (n-1)) where n is size of bucket.Could someone explain why modulo is not used which will also guarantee output range within bucket size.

答案1

得分: 2

& 运行速度更快,代价是仅适用于二的幂次。具体来说,如果 x 为非负整数且 n 为二的幂次,则 x & (n - 1) == x % n。对于哈希表,x & (n - 1) 也可以达到你想要的效果 — 即使 x 为负数,x & (n - 1) 也是如此,但对于 x % n 则不然。

这是完整且唯一的原因。

英文:

& runs faster, in exchange for working only for powers of two. (Specifically, x & (n - 1) == x % n if x is nonnegative and n is a power of two. x & (n - 1) also does what you want for a hash table -- even if x is negative, x & (n - 1) isn't -- unlike x % n.)

That's the complete and only reason.

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  • 本文由 发表于 2020年8月23日 12:12:01
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