Java Double isFinite() method

The isFinite() method of Java Double class returns true if the argument passed is a finite floating-point number, otherwise returns false(in case of NaN and infinity arguments).

Syntax

Parameters

d- This is the double-precision number to be tested.

Return value

The isFinite() method returns true when finite floating-point value is passed as argument, otherwise returns false.

Example 1

public class Double_isFiniteMethodExample1 {

    public static void main(String[] args) {

        double d1 = 99991.22;

        Double d2 = new Double(d1/0);

         if(Double.isFinite(d2)){

             System.out.println("Value returned is : " +Double.isFinite(d2));

         }

         else{

             System.out.println("Value returned is : "+Double.isFinite(d2));

         }

    }

}

Test it Now

Output:

Value returned is : false

Example 2

public class Double_isFiniteMethodExample2 {

    public static void main(String[] args) {

        // will return true

        Double d1 = new Double (0X7f800000);

        System.out.println("Returned result = "+ Double.isFinite(d1));

        //will return true

        Double d2 =new Double (76E-43);

        System.out.println("Returned result = "+ Double.isFinite(d2));

        //will return false

        Double d3 =new Double (d1/0);

        System.out.println("Returned result = "+ Double.isFinite(d3));

        //will return false

       Double d4 =new Double (-d1/0);

        System.out.println("Returned result = "+ Double.isFinite(d4));

    }

}

Test it Now

Output:

Returned result = true
Returned result = true
Returned result = false
Returned result = false

Next TopicJava Double

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