更快地实现Math.round？

Question

Are there any drawbacks to this code, which appears to be a faster (and correct) version of java.lang.Math.round ?

public static long round(double d) {

    if (d > 0) {
        return (long) (d + 0.5d);
    } else {
        return (long) (d - 0.5d);
    }
}

It takes advantage of the fact that, in Java, truncating to long rounds in to zero.

Answer 1

There are some special cases which the built in method handles, which your code does not handle. From the documentation:

• If the argument is NaN , the result is 0.
• If the argument is negative infinity or any value less than or equal to the value of Integer.MIN_VALUE , the result is equal to the value of Integer.MIN_VALUE .
• If the argument is positive infinity or any value greater than or equal to the value of Integer.MAX_VALUE , the result is equal to the value of Integer.MAX_VALUE .

Answer 2

I've been testing this, and there is one key potential drawback which has not yet been described here: You are changing the rounding tie-breaking method.

Math.round() implements the "round half up" rule, whereas your round() method implements the "round half away from zero" rule.

For example:

• Math.round(-0.5d) => 0L
• Your.round(-0.5d) => -1L

This may or may not be a problem for you, but you should understand that the above method is not a drop-in replacement for Math.round() , even after the NaN and infinity considerations already outlined.

Another relevant question: Rounding negative numbers in Java

As for the performance, there is no doubt that the above method is significantly faster than Math.round() - it runs in about 35% of the time for randomly generated positive and negative values. This can be a worthwhile optimisation when calling this method in a tight loop. It's even better (25% of the runtime) when given only positive values, possibly because of the CPU using branch prediction .

Math.round() is ultimately implemented by a native JNI call, which might be the cause of the performance difference. This Sun/Oracle bug suggests there might be pure-Java version in j6u22, but I can't see where, and indeed Math.round() in j6u23 performs similarly to j6u16 in my tests. I have not tested on other versions.

Answer 3

Yes; you're not accounting for underflow or overflow. Pragmatically speaking, this may not matter for your application.