Fastest most efficient way to determine decimal value is integer in Java

Question

Given a double variable named sizeValue and sizeValue contains something other than 0, what is the most efficient way to determine that sizeValue contains a value that is an integer?

Currently i'm using

sizeValue % 1 == 0

any other faster ways?

Answer 1

give a try to Math.ceil:

private static boolean isInt(double x) {
     return x == Math.ceil(x);
}

EDIT

I've done some benchmarks with the following methods:

private static boolean isInt1(double x) {
    return x == (int) x;
}

private static boolean isInt2(double x) {
    return x == Math.ceil(x);
}

private static boolean isInt3(double x) {
    return x % 1 == 0;
} 

isInt1 is the faster of them (on a sunjre 1.6)

Answer 2

I am not sure if it is any faster, but you could cast your double to an int and test for equality:

double d = 123.456;
boolean dIsAnInteger = (d == (int)d);

Answer 3

Warning: would you consider 0.9999999999999999 an integer? Probably not. But watch this:

double val = 0;
for(int i=0;i<10;i++)
    System.out.println(val+=0.1);

This prints out:

0.1
0.2
0.30000000000000004
0.4
0.5
0.6
0.7
0.7999999999999999
0.8999999999999999
0.9999999999999999

Which is problematic because at the end of the program val contains something that you think should be integer but is not.

So I'd make my method a bit slower (but more correct) by integrating a delta like JUnit does:

private static boolean isInt(double x, double delta) {
    double ceil = Math.ceil(x);
    return x-delta<ceil && x+delta>ceil;
}

and of course provide the same method with a sensible default:

private static boolean isInt(double x) {
    return isInt(x, 0.000000001);
}

Now isInt(val) finally returns true.