Method Overloading in Java with example

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, that 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. Number of parameters.
For example: This is a valid case of overloading

add(int, int)
add(int, int, int)

2. Data type of parameters.
For example:

add(int, int)
add(int, float)

3. Sequence of Data type of parameters.
For example:

add(int, float)
add(float, int)

Invalid case of method overloading:
When I say argument list, I am not talking about return type of the method, for example if two methods have same name, same parameters and have 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 method call to its definition happens at Compile time.

Method Overloading examples

As discussed in the beginning of this guide, method overloading is done by declaring same method with different parameters. The parameters must be different in either of these: number, sequence or types of parameters (or arguments). Lets see examples of each of these cases.

Argument list is also known as parameter list

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

This example shows how method overloading is done by having 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 number of parameters.

Lets 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 same. Both methods are having same number, data types and 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 number of arguments are different.

Case 4:

float mymethod(int a, float b)
float mymethod(float var1, int var2)

Result: Perfectly fine. Valid case of overloading. Sequence of the data types of parameters are different, 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 same. Even though return type of methods are different, it is not a valid case. Since return type of method doesn’t matter while overloading a method.

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 same name and argument list cannot be defined in a same class.