我应该如何映射长哈希值code为int（）？ [英] How should I map long to int in hashCode()?

问题描述

我有一个范围，有一个长，其值唯一标识特定对象在我的整个系统，就像一个GUID字段的对象。我已经重写的Object.Equals（）来使用这个ID进行比较，beause我想这与对象的拷贝工作。现在，我要重写 Object.hash code（）也是如此，这基本上意味着我的映射长来一些 INT 返回值。

I have a range of objects that have a long field whose value uniquely identifies a particular object across my entire system, much like a GUID. I have overriden Object.equals() to use this id for comparison, beause I want it to work with copies of the object. Now I want to override Object.hashCode(), too, which basically means mapping my long to some int return value.

如果我理解散code的目的正确的，它主要是用在哈希表，所以均匀分布是可取的。这意味着，只返回 ID％2 ^ 32 就足够了。是所有的，或者我应该知道的东西？

If I understood the purpose of hashCode correctly, it is mainly used in hash tables, so a uniform distribution would be desirable. This would mean, simply returning id % 2^32 would suffice. Is that all, or should I be aware of something else?

推荐答案

如何

Long.valueOf(guid).hashCode();

您可以返回，或者做当您创建的GUID和变量进行缓存。

You could return it, or do it when you create the guid and cache it in a variable.

纵观文档这样做：

(int)(this.longValue()^(this.longValue()>>>32))

这是一个不错的解决方案，因为它利用了Java库 - 总是更好地利用过的东西，已经被测试过

This is a decent solution since it makes use of the Java library - always better to leverage off of something that has been tested already.

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