Java Double Multiplication explanation?

Question

I recently tested something out that I heard using the following code

public static void main(String[] args) { double x = 4.35 * 100; System.out.println(x); }.

I am interested as to why this produces 434.99999999999994 rather than 435.0 . Thanks

There's like a zillion questions about floating point precision on SO already... — xlecoustillier, Jul 10 '13 at 13:30
Double values are imprecise, see http://stackoverflow.com/questions/16707397/whats-wrong-with-this-simple-double-calculation/16708068#16708068 — Richard Tingle, Jul 10 '13 at 13:30
4.35 is 100.010110011001100110011001(10011001)..... recurrring, i.e. not exactly representable in binary. You'll always have small errors in the last few decimal places — Richard Tingle, Jul 10 '13 at 13:32

score 17 · Accepted Answer · edited Jan 17 '21 at 07:37

17

When you type:

double x = 4.35;

x is not stored as-is. It is stored in a approaching form (probably 4.349999999 in this case).

If you want exact result, please use BigDecimal.

You can learn about the accuracy problems of floating-point technology.

edited Jan 17 '21 at 07:37

Basil Bourque

answered Jul 10 '13 at 13:31

Arnaud Denoyelle

Even with BigDecimal, I'm facing problems. Consider this,BigDecimal bd = new BigDecimal(4.35); System.out.println(bd.multiply(new BigDecimal(6))); – Toshik Langade Jan 17 '21 at 09:31

1 Answers1