# Why does Math.round(0.49999999999999994) return 1?

JavaFloating PointDoubleRounding## Java Problem Overview

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.

## Java Solutions

## Solution 1 - Java

**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;
}
```

_{}

_{
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
}

## Solution 2 - Java

This appears to be a known bug (Java"">http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=6430675">Java bug 6430675: Math.round has surprising behavior for 0x1.fffffffffffffp-2) which has been fixed in Java 7.

## Solution 3 - Java

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.

## Solution 4 - Java

I've got the same on JDK 1.6 32-bit, but on Java 7 64-bit I've got 0 for 0.49999999999999994 which rounded is 0 and the last line is not printed. It seems to be a VM issue, however, using floating points, you should expect the results to differ a bit on various environments (CPU, 32- or 64-bit mode).

And, when using `round`

or inverting matrices, etc., these *bits* can make a huge difference.

x64 output:

```
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 0
```

## Solution 5 - Java

The answer hereafter is an excerpt of an Oracle bug report 6430675 at. Visit the report for the full explanation.

**The methods {Math, StrictMath.round are operationally defined as**

```
(long)Math.floor(a + 0.5d)
```

for double arguments. While this definition usually works as expected, it gives the surprising result of 1, rather than 0, for 0x1.fffffffffffffp-2 (0.49999999999999994).

The value 0.49999999999999994 is the greatest floating-point value less than 0.5. As a hexadecimal floating-point literal its value is 0x1.fffffffffffffp-2, which is equal to (2 - 2^52) * 2^-2. == (0.5 - 2^54). Therefore, the exact value of the sum

```
(0.5 - 2^54) + 0.5
```

is 1 - 2^54. This is halfway between the two adjacent floating-point numbers (1 - 2^53) and 1. In the IEEE 754 arithmetic round to nearest even rounding mode used by Java, when a floating-point results is inexact, the closer of the two representable floating-point values which bracket the exact result must be returned; if both values are equally close, the one which its last bit zero is returned. In this case the correct return value from the add is 1, not the greatest value less than 1.

While the method is operating as defined, the behavior on this input is very surprising; the specification could be amended to something more like "Round to the closest long, rounding ties up," which would allow the behavior on this input to be changed.