为什么Java的Double.compare（double，double）实现了它的样子？ [英] Why is Java's Double.compare(double, double) implemented the way it is?

问题描述

我在查看 compare（double，double）在Java标准库（6）中。它读为：

I was looking at the implementation of compare(double, double) in the Java standard library (6). It reads:

public static int compare(double d1, double d2) {
    if (d1 < d2)
        return -1;       // Neither val is NaN, thisVal is smaller
    if (d1 > d2)
        return 1;        // Neither val is NaN, thisVal is larger

    long thisBits = Double.doubleToLongBits(d1);
    long anotherBits = Double.doubleToLongBits(d2);

    return (thisBits == anotherBits ?  0 : // Values are equal
            (thisBits < anotherBits ? -1 : // (-0.0, 0.0) or (!NaN, NaN)
             1));                          // (0.0, -0.0) or (NaN, !NaN)
}

编辑：Merits是一个非常糟糕的单词选择。我想知道这是如何工作的。

edit: "Merits" was a (very) bad choice of words. I wanted to know how this works.

推荐答案

@ Shoover的答案是正确的，但有一点比这更多。

@Shoover's answer is correct, but there is a bit more to it than this.

只要假设Java设计者决定实现 equals（...）和 double 与 == 具有相同的语义>实例。这意味着 equals（）对于包装的NaN将总是返回 false 。现在考虑如果您尝试在地图或集合中使用包装的NaN会发生什么。

Just suppose that the Java designers had decided to implement equals(...) and compare(...) with the same semantics as == on the wrapped double instances. This would mean that equals() would always return false for a wrapped NaN. Now consider what would happen if you tried to use a wrapped NaN in a Map or Collection.

List<Double> l = new ArrayList<Double>();
l.add(Double.NaN);
if (l.contains(Double.NaN)) {
    // this wont be executed.
}

Map<Object,String> m = new HashMap<Object,String>();
m.put(Double.NaN, "Hi mum");
if (m.get(Double.NaN) != null) {
    // this wont be executed.
}

没有什么意义！

因此，Java设计者决定（正确地IMO）对于我们今天这些Double方法的更复杂（但更直观）的定义。

So the Java designers decided (rightly IMO) on the more complicated (but more intuitive) definition for these Double methods that we have today.

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