Why Math.round(0.499999999999999917) rounds to 1 on Java 6

April 08, 2012

Overview

There are two types of error representation error and arithmetic rounding error which are common in floating point calculations. These two error combine in this simple example, Math.round(0.499999999999999917) rounds to 1 in Java 6.

Representation error

Floating point is a base 2 format, which means all number are represented as a sum of powers of 2. e.g. 6.25 is 2^2 + 2^1 + 2^-2. However, even simple numbers like 0.1 cannot be represented exactly. This becomes obvious when converting to BigDecimal as it will preserve the value actually represented without rounding.

new BigDecimal(0.1)= 
    0.1000000000000000055511151231257827021181583404541015625
BigDecimal.valueOf(0.1)= 0.1

Using the constructor obtains the value actually represented, using valueOf gives the same rounded value you would see if you printed the double

When a number is parsed, it is rounded to the closest represented value. This means that there is a number slightly less than 0.5 which will be rounded to 0.5 because it is the closest represented value.

The following does a brute force search for the smallest value which rounded becomes 1.0

public static final BigDecimal TWO = BigDecimal.valueOf(2);

public static void main(String... args) {
    int digits = 80;

    BigDecimal low = BigDecimal.ZERO;
    BigDecimal high = BigDecimal.ONE;

    for (int i = 0; i <= 10 * digits / 3; i++) {
        BigDecimal mid = low.add(high).divide(TWO, digits, RoundingMode.HALF_UP);
        if (mid.equals(low) || mid.equals(high))
            break;
        if (Math.round(Double.parseDouble(mid.toString())) > 0)
            high = mid;
        else
            low = mid;
    }

    System.out.println("Math.round(" + low + ") is " + 
            Math.round(Double.parseDouble(low.toString())));
    System.out.println("Math.round(" + high + ") is " + 
            Math.round(Double.parseDouble(high.toString())));
}

The Source Code

On Java 7 you get the following result.

Math.round(0.49999999999999997224442438437108648940920829772949218749999999999999999999999999) is 0
Math.round(0.49999999999999997224442438437108648940920829772949218750000000000000000000000000) is 1

What is surprising is that in Java 6 you get the follow.

Math.round(0.49999999999999991673327315311325946822762489318847656250000000000000000000000000) is 0
Math.round(0.49999999999999991673327315311325946822762489318847656250000000000000000000000001) is 1

Where do these numbers come from?

The Java 7 value is the mid point between 0.5 and the previous represent value. Above this mid point, the value is rounded to 0.5 when parsed.

The Java 6 value is the mid point between value value before 0.5 and the value before that.

Value 0.5 is 0.5
The previous value is 0.499999999999999944488848768742172978818416595458984375
... and the previous is 0.49999999999999988897769753748434595763683319091796875

The mid point between 0.5
 and 0.499999999999999944488848768742172978818416595458984375
 is 0.4999999999999999722444243843710864894092082977294921875

... and the mid point between 0.499999999999999944488848768742172978818416595458984375
 and 0.49999999999999988897769753748434595763683319091796875
 is 0.4999999999999999167332731531132594682276248931884765625

Why is the Java 6 value smaller

In the Java 6 Javadoc Math.round(double) is defined as

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

The problem with this definition is that 0.49999999999999994 + 0.5 has a rounding error which results in the value 1.0.

In the Java 7 Javadoc Math.round(double) it simply states

Returns the closest long to the argument, with ties rounding up.

So how does Java 7 fix this?

The source code for Java 7's Math.round looks like

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;
}

The result for the largest value less than 0.5 is hard coded.

So what is 0x1.fffffffffffffp-2?

It is a hexi-decimal presentation of the floating point value. It is rarely used, but it is precise as all values can be represented without error (to a limit of 53 bits).

Vanilla Java