Java - Double precision floating point

Question

Possible Duplicate:
how to fix double precision issue in java

I have a small piece of code like this:

double number1 = 6;
double number2 = 5.99;
double result = number1 - number2;

However, the result == 0.009999999999999787 instead of 0.01

I know it is the issue of IEEE 754 standard, but I don't understand why. Could you please explain it for me?

The closest `double` to 5.99 is 5.9900000000000002131628207280300557613372802734375. — Daniel Fischer, Dec 10 '12 at 22:08
You got floating point problems, I feel bad for you, son. I've got .99 problems but infinite precision ain't .01. — yshavit, Dec 10 '12 at 22:30

score 4 · Answer 1 · answered Dec 10 '12 at 22:07

4

This is because float point numbers cannot be exactly represented with in binary system with limited bits (not without precision loss)

See: http://en.wikipedia.org/wiki/Loss_of_significance

answered Dec 10 '12 at 22:07

sampson-chen

45,805
12
84
81

1

This is not loss of significance, it is inaccuracy of the actual representation. – user207421 Dec 10 '12 at 22:10

score 1 · Answer 2 · answered Dec 10 '12 at 22:08

1

Because there is no .01 in floating point numbers. The fractional bits are expressed as 1/root 2 so you can get something like .0125 or what you have there but there is not .01 in floating point numbers. If you need exact precision use integers instead.

answered Dec 10 '12 at 22:08

evanmcdonnal

46,131
16
104
115

3

If you need exact precision use BigDecimal http://docs.oracle.com/javase/1.5.0/docs/api/java/math/BigDecimal.html – KerrM Dec 10 '12 at 22:10
@KerrM Good call. I didn't know Java had a nice abstraction for you. – evanmcdonnal Dec 10 '12 at 22:12

Java - Double precision floating point

2 Answers2