为什么HashSet的内部实现会创建虚拟对象以在HashMap中作为值插入而不是插入空值？ [英] Why the internal implementation of HashSet creates dummy objects to insert as values in HashMap rather than inserting nulls?

问题描述

HashSet是使用HashMap实现的，当我们向HashSet添加任何e1时，如果e1不在集合中，它会在HashMap中添加（e1，new Object（））。我的问题是为什么他们插入新的Object（），当他们可以插入像（e1，null），这是更优化的方法，因为没有创建新的对象。在这里插入空值有什么缺点吗？

HashSet is implemented using HashMap and when we add anything say e1 to HashSet, internally it adds (e1,new Object()) in the HashMap if e1 was not present in the set. My question is why they are inserting new Object(), when they could have inserted like (e1,null), which is more optimized approach as no new Objects are created. Is there any downside to inserting nulls here?

推荐答案

A HashSet 不每次新密钥将添加到地图中时，都不会添加新的对象。它确实使用 Object ，但每次都使用相同的 Object 。此值在 HashSet 源代码中命名为 PRESENT 。

A HashSet doesn't add a new Object each time a new key is put into the map. It does use an Object, but it uses the same Object each time. This value is named PRESENT in the HashSet source code.

add 方法在内部 HashMap上调用 put（key，PRESENT） 。 remove 方法在内部 HashMap remove（key） >，但它必须返回 boolean ，指示密钥是否存在。如果将 null 存储为值，那么 HashSet 将需要调用 containsKey 首先，然后删除，以确定密钥是否存在 - 额外开销。在这里，只有一个 Object 的内存开销，这是非常小的。

The add method calls put(key, PRESENT) on the internal HashMap. The remove method calls remove(key) on the internal HashMap, but it must return a boolean indicating whether the key was present. If null were stored as the value, then the HashSet would need to call containsKey first, then remove, to determine if the key was present -- additional overhead. Here, there is only the memory overhead of one Object, which is quite minimal.

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