How can I handle precision error with float in Java?

Question

I'm wondering what the best way to fix precision errors is in Java. As you can see in the following example, there are precision errors:

class FloatTest
{
  public static void main(String[] args)
  {
    Float number1 = 1.89f;
    
    for(int i = 11; i < 800; i*=2)
    {
      System.out.println("loop value: " + i);
      System.out.println(i*number1);
      System.out.println("");
    }
  }
}

The result displayed is:

loop value: 11

20.789999

loop value: 22

41.579998

loop value: 44

83.159996

loop value: 88

166.31999

loop value: 176

332.63998

loop value: 352

665.27997

loop value: 704

1330.5599

Also, if someone can explain why it only does it starting at 11 and doubling the value every time. I think all other values (or many of them at least) displayed the correct result.

Problems like this have caused me headache in the past and I usually use number formatters or put them into a String.

Edit: As people have mentioned, I could use a double, but after trying it, it seems that 1.89 as a double times 792 still outputs an error (the output is 1496.8799999999999).

I guess I'll try the other solutions such as BigDecimal

user_10458281 · Answer

I never had a problem with simple arithmetic precision in either Basic, Visual Basic, FORTRAN, ALGOL or other "primitive" languages. It is beyond comprehension that JAVA can't do simple arithmetic without introducing errors. I need just two digits to the right of the decimal point for doing some accounting. Using Float subtracting 1000 from 1355.65 I get 355.650002! In order to get around this ridiculous error I have implemented a simple solution. I process my input by separating the values on each side of the decimal point as character, convert each to integers, multiply each by 1000 and add the two back together as integers. Ridiculous but there are no errors introduced by the poor JAVA algorithms.

How can I handle precision error with float in Java?

9 Answers