Java's ByteBuffer compareTo: Comparing a byte to itself "For float and double"?

Inside of Java's ByteBuffer is the method compareTo for implementing Comparable<ByteBuffer>...

public int compareTo(ByteBuffer that) {
    int n = this.position() + Math.min(this.remaining(), that.remaining());
    for (int i = this.position(), j = that.position(); i < n; i++, j++) {
        byte v1 = this.get(i);
        byte v2 = that.get(j);
        if (v1 == v2)
            continue;
        if ((v1 != v1) && (v2 != v2)) // For float and double
            continue;
        if (v1 < v2)
            return -1;
        return +1;
    }开发者_Go百科
    return this.remaining() - that.remaining();
}

What is the point of if ((v1 != v1) && (v2 != v2)) // For float and double?

If I write this kind of code, I'm warned about comparing identical expressions.

In that particular code, it doesn't make sense. However, that's a popular idiom for floats and doubles. Both can be "NaN", meaning they don't represent a valid, comparable number. NaN is technically not equal to anything, even itself, so for a float or double that code would check to see if both are NaN, which is enough in this case to call them equal.

I seems the Buffer classes are generated from a common source code template, this particular line only makes sense for the float and double versions to detect the special case where both numbers are NANs.

Double.NaN == Double.NaN and Float.NaN == Float.NaN gives false.

See also end of JLS 4.2.3:

The equality operator == returns false if either operand is NaN, and the inequality operator != returns true if either operand is NaN (§15.21.1).

As all the other anwsers already pointed out, it the check for Float.NaN and Double.NaN. They should have used Float.isNan and Double.isNaN. This would have made that code way more readable. Long story short: Don't emulate.