Java Double isFinite() Method with Examples

by Online Tutorials Library July 22, 2022

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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