How to get the decimal part of a float?

I need to extract the decimal part of a float number, but I get weird results:

float n = 22.65f;
// I want x = 0.65f, but...

x = n % 1; // x = 0.6499996

x = n - Math.floor(n); // x = 0.6499996185302734

x = n - (int)n; // x = 0.6499996

Why does this happen? Why do I get those values instead of 0.65?

Solution 1:

float only has a few digit of precision so you should expect to see a round error fairly easily. try double this has more accuracy but still has rounding errors. You have to round any answer you get to have a sane output.

If this is not desireable you can use BigDecimal which does not have rounding errors, but has its own headaches IMHO.

EDIT: You may find this interesting. The default Float.toString() uses minimal rounding, but often its not enough.

System.out.println("With no rounding");
float n = 22.65f;
System.out.println("n= "+new BigDecimal(n));
float expected = 0.65f;
System.out.println("expected= "+new BigDecimal(expected));

System.out.println("n % 1= "+new BigDecimal(n % 1));
System.out.println("n - Math.floor(n) = "+new BigDecimal(n - Math.floor(n)));
System.out.println("n - (int)n= "+new BigDecimal(n - (int)n));

System.out.println("With rounding");
System.out.printf("n %% 1= %.2f%n", n % 1);
System.out.printf("n - Math.floor(n) = %.2f%n", n - Math.floor(n));
System.out.printf("n - (int)n= %.2f%n", n - (int)n);

Prints

With no rounding
n= 22.6499996185302734375
expected= 0.64999997615814208984375
n % 1= 0.6499996185302734375
n - Math.floor(n) = 0.6499996185302734375
n - (int)n= 0.6499996185302734375
With rounding
n % 1= 0.65
n - Math.floor(n) = 0.65
n - (int)n= 0.65

Solution 2:

I think this would be the most simple way :

float n = 22.65f;
float x = n - (int) n;