如何从三个long生成哈希码 [英] How to generate a hash code from three longs

问题描述

我有一个坐标为键的HashMap。

I have a HashMap with coordinates as keys.

坐标有3个长整数，包含x，y和z坐标。 （坐标是并且需要是自定义类，坐标需要很长）。

Coordinates have 3 longs holding the x, y and z coordinate. (Coordinate is and needs to be a custom class, the coordinates need to be longs).

现在我希望能够访问例如字段[5,10,4]： hashMap.get（新坐标（5,10,4））。

Now i want to be able to access e.g. the field [5, 10, 4] by doing: hashMap.get(new Coordinate(5, 10, 4)).

我已经实现了equals方法，但这还不够，因为显然我还需要为hashCode提供一个实现。所以我的问题是如何从三个长度生成一个唯一的hashCode？。

I have implemented the equals method but that is not enough since apparently i need to provide an implementation for hashCode as well. So my question is how do i generate an unique hashCode from three longs?.

附加：使用外部库中的哈希生成器不是选项。

Additional: Using a hash generator from an external library is not option.

推荐答案

Joshua Bloch告诉您如何在第3章他的Effective Java。

Joshua Bloch tells you how to write equals and hashCode for your Coordinate class in chapter 3 of his "Effective Java".

像这样：

public class Coordinate
{
    private long x;
    private long y;
    private long z;

    @Override
    public boolean equals(Object o)
    {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;

        Coordinate that = (Coordinate) o;

        if (x != that.x) return false;
        if (y != that.y) return false;
        if (z != that.z) return false;

        return true;
    }

    @Override
    public int hashCode()
    {
        int result = (int) (x ^ (x >>> 32));
        result = 31 * result + (int) (y ^ (y >>> 32));
        result = 31 * result + (int) (z ^ (z >>> 32));
        return result;
    }
}

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