Given 开发者_StackOverflow社区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?
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)
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);
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.
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