这个HashSet如何产生排序输出？ [英] How is this HashSet producing sorted output?

问题描述

以下代码生成输出 [1,2] ，即使hashset未排序。

The following code produces the out put [1,2]even though hashset is not sorted.

Set set = new HashSet();
set.add(new Integer(2));
set.add(new Integer(1));
System.out.println(set);

为什么会这样？

推荐答案

此行为是由以下几个原因造成的：

This behaviour is caused by several separate reasons:

• 整数哈希自己


• 在Java中， HashMap s和 HashSet 由数组备份


• 他们还使用较高位修改哈希值来修改低位;如果散列在0..15的范围内，则不会被修改


• 对象的内容取决于修改散列的低位


• 当迭代地图或集合时，内部表将按顺序扫描

• Integers hash to themselves
• in Java, HashMaps and HashSets are backed up by an array
• they also modify hashes using the higher bits to modify the lower bits; if the hash is in range 0..15, it's therefore not modified
• what bucket an object goes depends on the lower bits of the modified hash
• when iterating over a map or a set, the inner table is scanned sequentially

因此，如果添加几个小（<16）对hashmap / hashset的整数，会发生这种情况：

So if you add several small (<16) integers to a hashmap/hashset, this is what happens:

• 整数 i 有hashcode i


• 因为它小于16，所以修改后的散列也是 i


• 它落在桶中。 i


• 在迭代时，会按顺序访问存储桶，因此如果存储的所有数据都是小整数，则会以升序检索它们订单

• integer i has hashcode i
• since it's less than 16, it's modified hash is also i
• it lands in the bucket no. i
• when iterating, the buckets are visited sequentially, so if all you stored there are small integers, they will be retrieved in ascending order

请注意，如果存储桶的初始数量太小，整数可能会落在以后编号的存储桶中：

Note that if the initial number of buckets is too small, the integers may land in buckets not numbered after them:

HashSet<Integer> set = new HashSet<>(4);
set.add(5); set.add(3); set.add(1);
for(int i : set) {
  System.out.print(i);
}

打印 153 。

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