comparison - Why is Java's Double.compare(double, double) implemented the way it is?

Question

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)
}

What are the merits of this implementation?

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

Answer 1

The explanation is in the comments in the code. Java has double values for both 0.0 and -0.0, as well as "not a number" (NaN). You can't use simple == operator for these values. Take a peek into the doubleToLongBits() source and at the Javadoc for the Double.equals() method:

Note that in most cases, for two instances of class Double, d1 and d2, the value of d1.equals(d2) is true if and only if

d1.doubleValue() == d2.doubleValue()

also has the value true. However, there are two exceptions:

• If d1 and d2 both represent Double.NaN, then the equals method returns true, even though Double.NaN == Double.NaN has the value false.
• If d1 represents +0.0 while d2 represents -0.0, or vice versa, the equal test has the value false, even though +0.0 == -0.0 has the value true.

This definition allows hash tables to operate properly.