How to resolve a Java Rounding Double issue [duplicate]

Answer 1

So far the most elegant and most efficient way to do that in Java:

double newNum = Math.floor(num * 100 + 0.5) / 100;

Answer 2

Better yet use JScience as BigDecimal is fairly limited (e.g., no sqrt function)

double dCommission = 1586.6 - 708.75;
System.out.println(dCommission);
> 877.8499999999999

Real dCommissionR = Real.valueOf(1586.6 - 708.75);
System.out.println(dCommissionR);
> 877.850000000000

Answer 3

This is a fun issue.

The idea behind Timons reply is you specify an epsilon which represents the smallest precision a legal double can be. If you know in your application that you will never need precision below 0.00000001 then what he suggests is sufficient to get a more precise result very close to the truth. Useful in applications where they know up front their maximum precision (for in instance finance for currency precisions, etc)

However the fundamental problem with trying to round it off is that when you divide by a factor to rescale it you actually introduce another possibility for precision problems. Any manipulation of doubles can introduce imprecision problems with varying frequency. Especially if you're trying to round at a very significant digit (so your operands are < 0) for instance if you run the following with Timons code:

System.out.println(round((1515476.0) * 0.00001) / 0.00001);

Will result in 1499999.9999999998 where the goal here is to round at the units of 500000 (i.e we want 1500000)

In fact the only way to be completely sure you've eliminated the imprecision is to go through a BigDecimal to scale off. e.g.

System.out.println(BigDecimal.valueOf(1515476.0).setScale(-5, RoundingMode.HALF_UP).doubleValue());

Using a mix of the epsilon strategy and the BigDecimal strategy will give you fine control over your precision. The idea being the epsilon gets you very close and then the BigDecimal will eliminate any imprecision caused by rescaling afterwards. Though using BigDecimal will reduce the expected performance of your application.

It has been pointed out to me that the final step of using BigDecimal to rescale it isn't always necessary for some uses cases when you can determine that there's no input value that the final division can reintroduce an error. Currently I don't know how to properly determine this so if anyone knows how then I'd be delighted to hear about it.

Answer 4

Any time you do calculations with doubles, this can happen. This code would give you 877.85:

double answer = Math.round(dCommission * 100000) / 100000.0;

Answer 5

Although you should not use doubles for precise calculations the following trick helped me if you are rounding the results anyway.

public static int round(Double i) {
    return (int) Math.round(i + ((i > 0.0) ? 0.00000001 : -0.00000001));
}

Example:

    Double foo = 0.0;
    for (int i = 1; i <= 150; i++) {
        foo += 0.00010;
    }
    System.out.println(foo);
    System.out.println(Math.round(foo * 100.0) / 100.0);
    System.out.println(round(foo*100.0) / 100.0);

Which prints:

0.014999999999999965
0.01
0.02

More info: http://en.wikipedia.org/wiki/Double_precision

Answer 6

To control the precision of floating point arithmetic, you should use java.math.BigDecimal. Read The need for BigDecimal by John Zukowski for more information.

Given your example, the last line would be as following using BigDecimal.

import java.math.BigDecimal;

BigDecimal premium = BigDecimal.valueOf("1586.6");
BigDecimal netToCompany = BigDecimal.valueOf("708.75");
BigDecimal commission = premium.subtract(netToCompany);
System.out.println(commission + " = " + premium + " - " + netToCompany);

This results in the following output.

877.85 = 1586.6 - 708.75