Floating point rounding (in java)

Question

What is the best way of determining if a given float(or double) has no significant decimal places.

f(234.0)  = true
f(34.45)  = false
f(3.1322) = false

ie equivalent of

EQ(((int)number) * 1.0 , number)

where EQ is a given method to compare floating points and it is OK to assume that the float fits in an integer.

Answer 1

Math.rint(x) == x

Math.rint() returns a double , so it also works for large numbers where the long result of Math.round() overflows.

Note that this also gives true for positive and negative infinity. You can explicitly exclude them by Math.rint(x) == x && !Double.isInfinite(x) .

Answer 2

Round the value to the nearest integer, and calculate the absolute difference to the actual value.

If that difference is less than a certain percentage of the actual value you are close "enough".

Answer 3

You could try something like this:

public static boolean f(double d) {
    return d % 1 == 0;
}