Retain precision with double in Java

Question

public class doublePrecision {
    public static void main(String[] args) {

        double total = 0;
        total += 5.6;
        total += 5.8;
        System.out         


        

Answer 1

You can't, because 7.3 doesn't have a finite representation in binary. The closest you can get is 2054767329987789/2**48 = 7.3+1/1407374883553280.

Take a look at http://docs.python.org/tutorial/floatingpoint.html for a further explanation. (It's on the Python website, but Java and C++ have the same "problem".)

The solution depends on what exactly your problem is:

• If it's that you just don't like seeing all those noise digits, then fix your string formatting. Don't display more than 15 significant digits (or 7 for float).
• If it's that the inexactness of your numbers is breaking things like "if" statements, then you should write if (abs(x - 7.3) < TOLERANCE) instead of if (x == 7.3).
• If you're working with money, then what you probably really want is decimal fixed point. Store an integer number of cents or whatever the smallest unit of your currency is.
• (VERY UNLIKELY) If you need more than 53 significant bits (15-16 significant digits) of precision, then use a high-precision floating-point type, like BigDecimal.