Checking if a number is an Integer in Java

Question

Is there any method or quick way to check whether a number is an Integer (belongs to Z field) in Java?

I thought of maybe subtracting it from the rounded number, but

Answer 1

    if((number%1)!=0)
    {
        System.out.println("not a integer");
    }
    else
    {
        System.out.println("integer");
    }

Answer 2

 int x = 3;

 if(ceil(x) == x) {

  System.out.println("x is an integer");

 } else {

  System.out.println("x is not an integer");

 }

Answer 3

One example more :)

double a = 1.00

if(floor(a) == a) {
   // a is an integer
} else {
   //a is not an integer.
}

In this example, ceil can be used and have the exact same effect.

Answer 4

Quick and dirty...

if (x == (int)x)
{
   ...
}

edit: This is assuming x is already in some other numeric form. If you're dealing with strings, look into Integer.parseInt.

Answer 5

if you're talking floating point values, you have to be very careful due to the nature of the format.

the best way that i know of doing this is deciding on some epsilon value, say, 0.000001f, and then doing something like this:

boolean nearZero(float f)
{
    return ((-episilon < f) && (f <epsilon)); 
}

then

if(nearZero(z-(int)z))
{ 
    //do stuff
}

essentially you're checking to see if z and the integer case of z have the same magnitude within some tolerance. This is necessary because floating are inherently imprecise.

NOTE, HOWEVER: this will probably break if your floats have magnitude greater than Integer.MAX_VALUE (2147483647), and you should be aware that it is by necessity impossible to check for integral-ness on floats above that value.

Answer 6

/**
 * Check if the passed argument is an integer value.
 *
 * @param number double
 * @return true if the passed argument is an integer value.
 */
boolean isInteger(double number) {
    return number % 1 == 0;// if the modulus(remainder of the division) of the argument(number) with 1 is 0 then return true otherwise false.
}