Method Overloading is a feature that allows a class to have more than one method having the same name if their argument lists are different. It is similar to constructor overloading in Java, which allows a class to have more than one constructor having different argument lists.
let’s get back to the point, when I say argument list it means the parameters that a method has: For example, the argument list of a method add(int a, int b) having two parameters is different from the argument list of the method add(int a, int b, int c) having three parameters.

Three ways to overload a method

In order to overload a method, the argument lists of the methods must differ in either of these:
1. A number of parameters.
For example, This is a valid case of overloading
add(int, int)
add(int, int, int)
2. The data type of parameters.
For example:
add(int, int)
add(int, float)
3. The sequence of Data type of parameters.
For example:
add(int, float)
add(float, int)
Invalid case of method overloading:
When I say the argument list, I am not talking about the return type of the method, for example, if two methods have the same name, same parameters and have a different return type, then this is not a valid method overloading example. This will throw compilation error.
int add(int, int)
float add(int, int)
Method overloading is an example of Static Polymorphism. We will discuss polymorphism and types of it in a separate tutorial.
Points to Note:
1. Static Polymorphism is also known as compile-time binding or early binding.
2. Static binding happens at compile time. Method overloading is an example of static binding where binding of the method call to its definition happens at Compile time.

Method Overloading examples

As discussed at the beginning of this guide, method overloading is done by declaring the same method with different parameters. The parameters must be different in either of these: number, sequence or types of parameters (or arguments). Let's see examples of each of these cases.
Argument list is also known as parameter list

Example 1: Overloading – Different Number of parameters in the argument list

This example shows how method overloading is done by having a different number of parameters
class DisplayOverloading
{
    public void disp(char c)
    {
         System.out.println(c);
    }
    public void disp(char c, int num)  
    {
         System.out.println(c + " "+num);
    }
}
class Sample
{
   public static void main(String args[])
   {
       DisplayOverloading obj = new DisplayOverloading();
       obj.disp('a');
       obj.disp('a',10);
   }
}
Output:
a
a 10
In the above example – method disp() is overloaded based on the number of parameters – We have two methods with the name disp but the parameters they have are different. Both are having different numbers of parameters.

Example 2: Overloading – Difference in the data type of parameters

In this example, method disp() is overloaded based on the data type of parameters – We have two methods with the name disp(), one with a parameter of char type and another method with the parameter of int type.
class DisplayOverloading2
{
    public void disp(char c)
    {
        System.out.println(c);
    }
    public void disp(int c)
    {
       System.out.println(c );
    }
}

class Sample2
{
    public static void main(String args[])
    {
        DisplayOverloading2 obj = new DisplayOverloading2();
        obj.disp('a');
        obj.disp(5);
    }
}
Output:
a
5

Example3: Overloading – Sequence of the data type of arguments

Here method disp() is overloaded based on the sequence of the data types of parameters – Both the methods have a different sequence of data types in the argument list. The first method is having an argument list as (char, int) and second is having (int, char). Since the sequence is different, the method can be overloaded without any issues.
class DisplayOverloading3
{
   public void disp(char c, int num)
   {
       System.out.println("I’m the first definition of method disp");
   }
   public void disp(int num, char c)
   {
       System.out.println("I’m the second definition of method disp" );
   }
}
class Sample3
{
   public static void main(String args[])
   {
       DisplayOverloading3 obj = new DisplayOverloading3();
       obj.disp('x', 51 );
       obj.disp(52, 'y');
   }
}
Output:
I’m the first definition of method disp
I’m the second definition of method disp

Method Overloading and Type Promotion

When a data type of smaller size is promoted to the data type of bigger size than this is called type promotion, for example, the byte data type can be promoted to short, a short data type can be promoted to int, long, double, etc.
What it has to do with method overloading?
Well, it is very important to understand type promotion else you will think that the program will throw compilation error but in fact that program will run fine because of type promotion.
Let's take an example to see what I am talking here:
class Demo{
   void disp(int a, double b){
 System.out.println("Method A");
   }
   void disp(int a, double b, double c){
 System.out.println("Method B");
   }
   public static void main(String args[]){
 Demo obj = new Demo();
 /* I am passing float value as a second argument but
  * it got promoted to the type double, because there
  * wasn't any method having arg list as (int, float)
  */
 obj.disp(100, 20.67f);
   }
}
Output:
Method A
As you can see that I have passed the float value while calling the disp() method but it got promoted to the double type as there wasn’t any method with argument list as (int, float)
But this type of promotion doesn’t always happen, let's see another example:
class Demo{
   void disp(int a, double b){
 System.out.println("Method A");
   }
   void disp(int a, double b, double c){
 System.out.println("Method B");
   }
   void disp(int a, float b){
 System.out.println("Method C");
   }
   public static void main(String args[]){
 Demo obj = new Demo();
 /* This time promotion won't happen as there is
  * a method with arg list as (int, float)
  */
 obj.disp(100, 20.67f);
   }
}
Output:
Method C
As you see that this time type promotion didn’t happen because there was a method with matching argument type.
Type Promotion table:
The data type on the left side can be promoted to any of the data types present on the right side of it.
byte → short → int → long
short → int → long
int → long → float → double
float → double
long → float → double

Let's see few Valid/invalid cases of method overloading

Case 1:
int mymethod(int a, int b, float c)
int mymethod(int var1, int var2, float var3)
Result: Compile time error. Argument lists are exactly the same. Both methods are having the same number, data types and the same sequence of data types.
Case 2:
int mymethod(int a, int b)
int mymethod(float var1, float var2)
Result: Perfectly fine. Valid case of overloading. Here data types of arguments are different.
Case 3:
int mymethod(int a, int b)
int mymethod(int num)
Result: Perfectly fine. Valid case of overloading. Here the number of arguments is different.
Case 4:
float mymethod(int a, float b)
float mymethod(float var1, int var2)
Result: Perfectly fine. Valid case of overloading. The sequence of the data types of parameters is different, the first method is having (int, float) and second is having (float, int).
Case 5:
int mymethod(int a, int b)
float mymethod(int var1, int var2)
Result: Compile time error. Argument lists are exactly the same. Even though the return type of methods is different, it is not a valid case. Since return type of method doesn’t matter while overloading a method.
Guess the answers before checking it at the end of programs:
Question 1 – return type, method name and argument list same.
class Demo
{
   public int myMethod(int num1, int num2)
   { 
       System.out.println("First myMethod of class Demo");
       return num1+num2;
   }
   public int myMethod(int var1, int var2)
   {
       System.out.println("Second myMethod of class Demo");
       return var1-var2;
   }
}
class Sample4
{
   public static void main(String args[])
   {
       Demo obj1= new Demo();
       obj1.myMethod(10,10);
       obj1.myMethod(20,12);
   }
}
Answer:
It will throw a compilation error: More than one method with the same name and argument list cannot be defined in the same class.
Question 2 – the return type is different. Method name & argument list the same.
class Demo2
{
   public double myMethod(int num1, int num2)
   {
      System.out.println("First myMethod of class Demo");
      return num1+num2;
   }
   public int myMethod(int var1, int var2)
   {
      System.out.println("Second myMethod of class Demo");
      return var1-var2;
   }
}
class Sample5
{
   public static void main(String args[])
   {
      Demo2 obj2= new Demo2();
      obj2.myMethod(10,10);
      obj2.myMethod(20,12);
   }
}
Answer:
It will throw a compilation error: More than one method with the same name and argument list cannot be given in a class even though their return type is different. The method return type doesn’t matter in case of overloading.

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