为什么 HashMap 重新散列键对象提供的哈希码? [英] Why does a HashMap rehash the hashcode supplied by the key object?

问题描述

我正在阅读 Java 1.6 API 提供的 HashMap 类的代码，无法完全理解以下操作的需要(在 put 和 get 方法的主体中找到):

I am reading the code of the HashMap class provided by the Java 1.6 API and unable to fully understand the need of the following operation (found in the body of put and get methods):

int hash = hash(key.hashCode());

其中方法 hash() 具有以下主体:

where the method hash() has the following body:

 private static int hash(int h) {
         h ^= (h >>> 20) ^ (h >>> 12);
    return h ^ (h >>> 7) ^ (h >>> 4);
}

这通过对提供的哈希码执行位操作来有效地重新计算哈希.即使 API 如下所述，我也无法理解这样做的必要性:

This effectively recalculates the hash by executing bit operations on the supplied hashcode. I'm unable to understand the need to do so even though the API states it as follows:

这很关键因为 HashMap 使用二次幂长度的哈希表，所以否则会遇到相同的 hashCode 冲突低位.

This is critical because HashMap uses power-of-two length hash tables, that otherwise encounter collisions for hashCodes that do not differ in lower bits.

我确实理解键值解析存储在数据结构数组中，并且该数组中项目的索引位置由其哈希决定.我不明白的是这个函数如何向散列分布添加任何值.

I do understand that the key value pars are stored in an array of data structures, and that the index location of an item in this array is determined by its hash. What I fail to understand is how would this function add any value to the hash distribution.

推荐答案

正如 Helper 所写，它只是为了防止关键对象的现有散列函数出现故障并且不能很好地混合较低的位.根据pgras引用的来源，

As Helper wrote, it is there just in case the existing hash function for the key objects is faulty and does not do a good-enough job of mixing the lower bits. According to the source quoted by pgras,

 /**
  * Returns index for hash code h.
  */
 static int indexFor(int h, int length) {
     return h & (length-1);
 }

散列以 2 的幂的长度进行 AND 运算(因此，length-1 保证是 1 的序列).由于这种与运算，只有 h 的低位被使用.h 的其余部分被忽略.想象一下，无论出于何种原因，原始散列只返回可被 2 整除的数字.如果直接使用它，则永远不会使用散列图的奇数位置，导致冲突次数增加 x2.在真正病态的情况下，错误的散列函数会使散列图表现得更像一个列表，而不是一个 O(1) 容器.

The hash is being ANDed in with a power-of-two length (therefore, length-1 is guaranteed to be a sequence of 1s). Due to this ANDing, only the lower bits of h are being used. The rest of h is ignored. Imagine that, for whatever reason, the original hash only returns numbers divisible by 2. If you used it directly, the odd-numbered positions of the hashmap would never be used, leading to a x2 increase in the number of collisions. In a truly pathological case, a bad hash function can make a hashmap behave more like a list than like an O(1) container.

Sun 工程师必须运行测试表明，太多哈希函数的低位不够随机，并且许多哈希图不够大，无法使用高位.在这些情况下，HashMap 的 hash(int h) 中的位操作可以提供比大多数预期用例(由于较低的冲突率)的净改进，即使需要额外的计算.

Sun engineers must have run tests that show that too many hash functions are not random enough in their lower bits, and that many hashmaps are not large enough to ever use the higher bits. Under these circumstances, the bit operations in HashMap's hash(int h) can provide a net improvement over most expected use-cases (due to lower collision rates), even though extra computation is required.

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