Java.lang.Double.compareTo() Method


The java.lang.Double.compareTo() method compares two Double objects numerically. There are two ways in which comparisons performed by this method differ from those performed by the Java language numerical comparison operators (<, <=, ==, >= >) when applied to primitive double values −

  • Double.NaN is considered by this method to be equal to itself and greater than all other double values (including Double.POSITIVE_INFINITY).
  • 0.0d is considered by this method to be greater than -0.0d.


Following is the declaration for java.lang.Double.compareTo() method

public int compareTo(Double anotherDouble)


anotherDouble − This is the Double to be compared.

Return Value

This method returns the value 0 if anotherDouble is numerically equal to this Double; a value less than 0 if this Double is numerically less than anotherDouble; and a value greater than 0 if this Double is numerically greater than anotherDouble.




The following example shows the usage of java.lang.Double.compareTo() method.

package com.tutorialspoint;

import java.lang.*;

public class DoubleDemo {

   public static void main(String[] args) {

      // compares two Double objects numerically
      Double obj1 = new Double("8.5");
      Double obj2 = new Double("11.50");
      int retval =  obj1.compareTo(obj2);
      if(retval > 0) {
         System.out.println("obj1 is greater than obj2");
      } else if(retval < 0) {
         System.out.println("obj1 is less than obj2");
      } else {
         System.out.println("obj1 is equal to obj2");

Let us compile and run the above program, this will produce the following result −

obj1 is less than obj2