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

When you input a double number, for example, 33.33333333333333, the value you get is actually the closest representable double-precision value, which is exactly:

33.3333333333333285963817615993320941925048828125

Dividing that by 100 gives:

0.333333333333333285963817615993320941925048828125

which also isn't representable as a double-precision number, so again it is rounded to the nearest representable value, which is exactly:

0.3333333333333332593184650249895639717578887939453125

When you print this value out, it gets rounded yet again to 17 decimal digits, giving:

0.33333333333333326