Method Overloading in Java

Method Overloading in Java

If a class have multiple methods by same name but different parameters, it is known as Method Overloading.

If we have to perform only one operation, having the same name of the methods increases the readabilityof the program. Suppose you have to perform addition of the given numbers but there can be any number of arguments, if you write the method such as a(int,int) for two parameters, and b(int,int,int) for three parameters then it may be difficult for you as well as other programmers to understand the behaviour of the method because its name differs. So, we perform method overloading to figure out the program quickly.


Advantage of method overloading?

Method overloading increases the readability of the program.

Different ways to overload the method:
There are two ways to overload the method in java

1. By changing number of arguments
2. By changing the data type

Note: In java, Methood Overloading is not possible by changing return type of the method.
1)Example of Method Overloading by changing the no. of arguments
In this example, we have created two overloaded methods, first sum method performs addition of two
numbers and second sum method performs addition of three numbers.
//Program of method overloading by chainging no. of arguments

class Calculation{
void sum(int a,int b){System.out.println(a+b);}
void sum(int a,int b,int c){System.out.println(a+b+c);}
public static void main(String args[]){
Calculation obj=new Calculation();
obj.sum(10,10,10);
obj.sum(20,20);
}
}
Output:30
40
2

2)Example of Method Overloading by changing data type of argument
In this example, we have created two overloaded methods that differs in data type. The first sum method
receives two integer arguments and second sum method receives two double arguments.
//Program of method overloading by chainging data type of argument

class Calculation{
void sum(int a,int b){System.out.println(a+b);}
void sum(double a,double b){System.out.println(a+b);}
public static void main(String args[]){
Calculation obj=new Calculation();
obj.sum(10.5,10.5);
obj.sum(20,20);
}
}

Output:21.0
40


Que) Why Method Overloaing is not possible by changing the return type of method?
In java, method overloading is not possible by changing the return type of the method because there may
occur ambiguity. Let's see how ambiguity may occur:
because there was problem:

class Calculation{
int sum(int a,int b){System.out.println(a+b);}
double sum(int a,int b){System.out.println(a+b);}
public static void main(String args[]){
Calculation obj=new Calculation();
int result=obj.sum(20,20); //Compile Time Error
}
}

int result=obj.sum(20,20); //Here how can java determine which sum() method should be called
 
Can we overload main() method?
Yes, by method overloading. You can have any number of main methods in a class by method overloading.
Let's see the simple example:
//Program of overloading main() method

class Simple{
public static void main(int a){
System.out.println(a);
}
public static void main(String args[]){
System.out.println("main() method invoked");
main(10);
}
}

Output:main() method invoked
10

Method Overloading and TypePromotion
One type is promoted to another implicitly if no matching datatype is found. Let's understand the concept by
the figure given below:



As displayed in the above diagram, byte can be promoted to short, int, long, float or double. The short
datatype can be promoted to int,long,float or double. The char datatype can be promoted to int,long,float or
double and so on.

Example of Method Overloading with TypePromotion
//Program of method overloading with TypePromotion
class Calculation{
void sum(int a,long b){System.out.println(a+b);}
void sum(int a,int b,int c){System.out.println(a+b+c);}
public static void main(String args[]){
Calculation obj=new Calculation();
obj.sum(20,20);//now second int literal will be promoted to long
obj.sum(20,20,20);
}
}
Output:40
60

Example of Method Overloading with TypePromotion if matching found
If there are matching type arguments in the method, type promotion is not performed.
//Program of method overloading with TypePromotion if matching found

class Calculation{
void sum(int a,int b){System.out.println("int arg method invoked");}
void sum(long a,long b){System.out.println("long arg method invoked");}
public static void main(String args[]){
Calculation obj=new Calculation();
obj.sum(20,20);//now int arg sum() method gets invoked
}
}
Output:int arg method invoked

Example of Method Overloading with TypePromotion in case ambiguity
If there are no matching type arguments in the method, and each method promotes similar number of
arguments, there will be ambiguity.


//Program of method overloading with TypePromotion in case ambiguity

class Calculation{
void sum(int a,long b){System.out.println("a method invoked");}
void sum(long a,int b){System.out.println("b method invoked");}
public static void main(String args[]){
Calculation obj=new Calculation();
obj.sum(20,20);//now ambiguity
}
}

Output:Compile Time Error