Method overloading in Java programming language

In this tutorial, we discuss Method overloading in Java programming language.

The class has multiple methods with the same name but distinguish parameter in the parameter list. It is called as method overloading in Java.

Method overloading is an Oop concept and is one of the features in Java language. It helps to have many methods with the same name but with some different techniques in the parameter list.

There are three different parameter list

Number of parameters
The data type of parameter
Order of data type in the parameter

When one or more method names are the same, we can use the different number of the parameter with the same data type.

Method overloading in Java programming language — Method overloading

Different number of parameter

add(int , int)  // method having two parameter as same name
add(int , int, int ) // method having three parameter as same name

2. The different data type of parameter

mul(int , int )
mul(int ,float ) //methods having two same number of parameter  different data type

3. Different order of data type in the parameter

div(int , float )
div(float, int ) //methods having two same number of parameter  different order

Different number of parameter in the argument list

This example shows How method overloading is performed by existing different number of parameter in the argument list

Example

Program 1

class OverloadingEx1{
public static void add(int a, int b,int c){
System.out.println("Total is :"+(a+b+c));

}
public static void add(int a, int b){//method 1
System.out.println("Total is :"+(a+b));

}
public static void main(String args[]){//method 2
add(45,67);//call method 1
add(45,67,34);//call method 2

}

When the above code is executed, it produces the following results:

Total is : 112
Total is : 146

In the above example, The class have two methods with the name add() but both are having the different number of parameter.

add(int a, int b,int c)
add(int a, int b)

So method add() is overloaded based in the different number of parameter

Program 2

class Overloading2{ //class_one
public static void add(float a, float b){,//method 1
System.out.println("Total is :"+(a+b));    
}
public static void add(float a, float b, float c){//method 2
System.out.println("Total is :"+(a+b+c));
}

}
class OverloadingEx2{//class_two
public static void main(String args[]){//main method
 Overloading2 sum=new Overloading2();  //create object
 sum.add(34.678f,23.00f);              //call method 1
 sum.add(26.958f,37.01f,98.011f);      //call method 1
}

}

When the above code is executed, it produces the following results:

Total is :57.678
Total is :161.979

Difference Data type of parameter in the argument list

This example shows How method overloading is performed by the existing different data type of parameter in the argument list

Example

Program 1

class OverloadingEx3{

public static void add(int a, int b){
System.out.println("Sum of a and b is :"+(a+b));
}
public static void add(String fname, String lname){
System.out.println("Full name is :"+fname+lname);
}

public static void main(String args[]){
    add(34,67);
    add("Robert ","Sam");
}
}

When the above code is executed, it produces the following results:

Sum of a and b is : 101
Full name is : Robert Sam

In the above example, The class have two methods with the name add() but both are having the different type of parameter.

add(int a, int b)
add(String name, String name)

So method add() is overloaded based in the different type of parameter in the argument list

3. Change Order of data type in the parameter

This example shows How method overloading is performed by existing different order of parameter in the argument list

Program 1

class OverloadingEx4{

public static void method_One(int a, String b){
System.out.println("This is a method one for overloading");
}
public static void method_One(String b, int a){
System.out.println("This is a method two for overloading");
}

public static void main(String args[]){
    method_One(34,"String");
    method_One("String",34);
}
}

When the above code is executed, it produces the following results:

This is a method one for overloading
This is a method two for overloading

In the above example, The class have two methods with the name method_One() but both are having a different order of parameter.

method_One(int a, String b)
method_One(String b, int a)

So method add() is overloaded based in different order of parameter in the argument list

Program 2

class OverloadingEx4{

public static void add(int a, float b){
System.out.println("Sum of a and b : "+(a+b));
}
public static void add(float b, int a){
System.out.println("Sum of a and b : "+(a+b));
}

public static void main(String args[]){
    add(34,47.76f);
    add(37.35f,78);
}
}

When the above code is executed, it produces the following results:

Sum of a and b : 81.759995
Sum of a and b : 115.35

Over view

invalid cases of method overloading

case 1

int add(int a,int b)
int add(int c,int d) //same method name and same parameter list

This format is not supported method overloading

Example

//example for return type ,method name and argument list same
class MethodOver1{
public int method1(int count1, int count2){
System.out.println(count1+count2);
return count1+count2;
}

public int method1(int var1, int var2){
System.out.println(count1+count2);
return count1+count2;

}
}
class disp{
    public static void main(String args[]){
        MetodOver1 obj=new MetodOver1();
        obj.metod1(45,67);
        obj.metod1(56,66);
    }
    
}

When the above code is executed, it produces the following results:

compile time error

case 2

int add(int a,int b)

float add(int c,int d) //same method name and same parameter list but different return type

This format is not supported method overloading

//example for different return type ,method name and argument list are same
class MethodOver2{
public int method1(int count1, int count2){//return type int
System.out.println(count1+count2);
return count1+count2;
}

public double method1(int var1, int var2){//return type double 
System.out.println(count1+count2);
return count1+count2;

}
}
class disp{
    public static void main(String args[]){
        MetodOver1 obj=new MetodOver1();
        obj.metod1(45,67);
        obj.metod1(56,66);
    }
    
    
    
}

When the above code is executed, it produces the following results:

compile time error

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