Overloading Methods

UNIT II : Inheritance, Packages and Interfaces

Overloading Methods

Overloading is a mechanism in which we can use many methods having the same function name but can pass different number of parameters or different types of parameter.

For example:

int sum(int a,int b);

double sum(double a, double b);

int sum(int a,int b,int c);

That means, by overloading mechanism, we can handle different number of parameters or different types of parameter by having the same method name.

Following Java program explains the concept overloading -

Java Program [OverloadingDemo.java]

public class OverloadingDemo {

public static void main(String args[]) {

System.out.println("Sum of two integers");

Sum(10,20); <-------------- line A

System.out.println("Sum of two double numbers");

Sum(10.5,20.4); <------------ line B

System.out.println("Sum of three integers");

Sum(10,20,30); <‒‒‒‒‒‒‒‒‒‒‒ line C

}

public static void Sum(int num1,int num2)

{

int ans;

ans num1+num2;

System.out.println(ans);

}

public static void Sum(double num1,double num2)

{

double ans;

ans num1+num2;

System.out.println(ans);

}

public static void Sum(int num1,int num2,int num3)

{

int ans;

ans num1+num2+num3;

System.out.println(ans)

}

}

Output

F:\test>javac OverloadingDemo.java

F:\test>java OverloadingDemo

Sum of two integers

30

Sum of two double numbers

30.9

Sum of three integers

60

Program Explanation

In above program, we have used three different methods possessing the same name. Note that,

on line A

We have invoked a method to which two integer parameters are passed. Then compiler automatically selects the definition public static void Sum(double num1,double num2) to fulfill the call. Hence we get the output as 30 which is actually the addition of 10 and 20.

on line B

We have invoked a method to which two double parameters are passed. Then compiler automatically selects the definition public static void Sum(double num1,double num2) to fulfill the call. Hence we get the output as 30.9 which is actually the addition of 10.5 and 20.4.

on line C

We have invoked a method to which the three integer parameters are passed. Then compiler automatically selects the definition public static void Sum(int num1,int num2,int num3) to fulfill the call. Hence we get the output as 60 which is actually the addition of 10, 20 and 30.