为什么 Java 的 Double.compare(double, double) 是这样实现

Why is Java#39;s Double.compare(double, double) implemented the way it is?(为什么 Java 的 Double.compare(double, double) 是这样实现的?)

问题描述

我正在查看 compare(double, double) 在 Java 标准库 (6) 中.上面写着:

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)
}

这个实现的优点是什么?

What are the merits of this implementation?

优点"是一个(非常)糟糕的词选择.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(阅读！)，但除此之外还有更多内容.

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

作为 javadoc for Double::equals 状态:

As the javadoc for Double::equals states:

这个定义允许哈希表正常运行."

"This definition allows hash tables to operate properly."

假设 Java 设计者决定用与 ==<相同的语义来实现 equals(...) 和 compare(...)/code> 在包装的 double 实例上.这意味着 equals() 将始终为包装的 NaN 返回 false.现在考虑如果您尝试在 Map 或 Collection 中使用包装的 NaN 会发生什么.

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.
}