Why does Math.round(0.49999999999999994) return 1?
In the following program you can see that each value slightly less than .5 is rounded down, except for 0.5 .
for (int i = 10; i >= 0; i--) {
long l = Double.doubleToLongBits(i + 0.5);
double x;
do {
x = Double.longBitsToDouble(l);
System.out.println(x + " rounded is " + Math.round(x));
l--;
} while (Math.round(x) > i);
}
prints
10.5 rounded is 11
10.499999999999998 rounded is 10
9.5 rounded is 10
9.499999999999998 rounded is 9
8.5 rounded is 9
8.499999999999998 rounded is 8
7.5 rounded is 8
7.499999999999999 rounded is 7
6.5 rounded is 7
6.499999999999999 rounded is 6
5.5 rounded is 6
5.499999999999999 rounded is 5
4.5 rounded is 5
4.499999999999999 rounded is 4
3.5 rounded is 4
3.4999999999999996 rounded is 3
2.5 rounded is 3
2.4999999999999996 rounded is 2
1.5 rounded is 2
1.4999999999999998 rounded is 1
0.5 rounded is 1
0.49999999999999994 rounded is 1
0.4999999999999999 rounded is 0
I am using Java 6 update 31.
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("%016xn", 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
这似乎是一个已知错误（Java bug 6430675：Math.round对于0x1.fffffffffffffp-2具有令人惊讶的行为），这在Java 7中已得到修复。
Source code in JDK 6:
public static long round(double a) {
return (long)Math.floor(a + 0.5d);
}
Source code in JDK 7:
public static long round(double a) {
if (a != 0x1.fffffffffffffp-2) {
// a is not the greatest double value less than 0.5
return (long)Math.floor(a + 0.5d);
} else {
return 0;
}
}
When the value is 0.49999999999999994d, in JDK 6, it will call floor and hence returns 1, but in JDK 7, the if condition is checking whether the number is the greatest double value less than 0.5 or not. As in this case the number is not the greatest double value less than 0.5, so the else block returns 0.
You can try 0.49999999999999999d, which will return 1, but not 0, because this is the greatest double value less than 0.5.
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