Weird Java math execution

Question

I have an weird issue. I'm trying to store the result of an equation into a double variable.

double s = (((100 + 1)*(1/3))/100 + (1/3));

This returns a value a 0 rather than .67 (the correct value calculated from a calculator). Any reason why this could happen?

Note: A solution of saying that I could just make s = .67 is not a solution,

Thanks in advance.

If you add `d` after all your numbers, eg `100d`, you'll use floating point operations, not integer ops. — jedwards, Mar 24 '13 at 14:21
As it is already answered, I am just adding a side-note that if you want precise value of `0.67` unlike something like `0.669999999999` you should use `BigDecimal`. http://stackoverflow.com/questions/322749/retain-precision-with-doubles-in-java — gaganbm, Mar 25 '13 at 13:57

score 8 · Accepted Answer · answered Mar 24 '13 at 14:19

8

The following uses integer (i.e. truncating) division, the result of which is zero:

1/3

To get floating-point division, turn either of the argument into a double, e.g.

1.0/3

Thus, the overall expression becomes:

double s = (((100 + 1)*(1./3))/100 + (1./3));

1. is the same as 1.0. Other ways to express the same number as a double are 1d and 1D.

The above expression evaluates to 0.6699999999999999.

answered Mar 24 '13 at 14:19

NPE

Yes, I don't know how I let this slip. Thanks so much. It's been bugging me for quite some time. – Stephen D Mar 24 '13 at 14:21

score 3 · Answer 2 · answered Mar 24 '13 at 14:22

3

The compiler sees your numers as int's..

try like this:

double s = (((100d + 1d)*(1d/3d))/100d + (1d/3d));

now the result will be:

0.6699999999999999

answered Mar 24 '13 at 14:22

Frank

2 Answers2