Why does Math.round(0.49999999999999994) return 1?

Summary

In Java 6 (and presumably earlier), round(x) is implemented as floor(x+0.5).¹ This is a specification bug, for precisely this one pathological case.² Java 7 no longer mandates this broken implementation.³

The problem

0.5+0.49999999999999994 is exactly 1 in double precision:

static void print(double d) {
    System.out.printf("%016x\n", Double.doubleToLongBits(d));
}

public static void main(String args[]) {
    double a = 0.5;
    double b = 0.49999999999999994;

    print(a);      // 3fe0000000000000
    print(b);      // 3fdfffffffffffff
    print(a+b);    // 3ff0000000000000
    print(1.0);    // 3ff0000000000000
}

This is because 0.49999999999999994 has a smaller exponent than 0.5, so when they’re added, its mantissa is shifted, and the ULP gets bigger.

The solution

Since Java 7, OpenJDK (for example) implements it thus:⁴

public static long round(double a) {
    if (a != 0x1.fffffffffffffp-2) // greatest double value less than 0.5
        return (long)floor(a + 0.5d);
    else
        return 0;
}

---

_{1. http://docs.oracle.com/javase/6/docs/api/java/lang/Math.html#round%28double%29}

_{2. http://bugs.java.com/bugdatabase/view_bug.do?bug_id=6430675 (credits to @SimonNickerson for finding this)}

_{3. http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#round%28double%29}

_{4. http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/7u40-b43/java/lang/Math.java#Math.round%28double%29}