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Method Overloading in Java with examples

By Chaitanya Singh | Filed Under: OOPs Concept

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.

Example 2: Overloading – Difference in 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 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 data type of arguments

Here method disp() is overloaded based on sequence of data type of parameters – Both the methods have different sequence of data type in argument list. First method is having 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: 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.
Lets 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 promotion doesn’t always happen, lets 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 the any of the data type present in the right side of it.

byte → short → int → long
short → int → long
int → long → float → double
float → double
long → float → double

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.

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

Question 2 – return type is different. Method name & argument list 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 same name and argument list cannot be given in a class even though their return type is different. Method return type doesn’t matter in case of overloading.

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Comments

  1. sirisha says

    August 27, 2014 at 3:56 PM

    Very clear. Excellent examples. Thanks a lot

    Reply
  2. sweety says

    October 13, 2014 at 6:38 PM

    Thanks.. very good and easy to understand

    Reply
  3. Syed Munir Uddin says

    October 23, 2014 at 1:42 PM

    Excellent Example keep it up and thanks :-)

    Reply
  4. vishal says

    November 13, 2014 at 3:45 PM

    very clear….

    Reply
  5. prerana says

    November 13, 2014 at 9:40 PM

    The examples helped very much. The examples were clear, and to the point. Great explanation. :D

    Reply
  6. Mayur says

    January 4, 2015 at 9:12 AM

    Very well presentation. Thanks a lot.

    Reply
  7. Leesa says

    January 28, 2015 at 3:56 PM

    The case 4-
    float mymethod(int a, float b)
    float mymethod(float var1, int var2)
    causes a compilation error due to implicit method invocation datatype conversion. Please check ..

    Reply
    • Shipra says

      June 15, 2016 at 11:27 AM

      that’s right

      Reply
    • hayenadeblue says

      November 23, 2016 at 12:30 PM

      I think what is written in the article is correct. Don’t understand why Shipra agrees.

      Reply
      • Rajan Gupta says

        February 8, 2017 at 7:05 PM

        This is throwing compilation error because you might be passing integer values in both the methods…if you specifically pass float then this will not throw compilation error
        Something like this:
        mymethod(1.0f, 10);
        mymethod(1, 10.0f);
        I hope this helps.

        Reply
  8. Noah says

    March 6, 2015 at 7:52 PM

    There couldn’t be a more detailed and useful explanation of method overloading than this.

    Thank you!

    Reply
  9. chetan says

    March 10, 2015 at 5:41 AM

    You have described in well manner i like that your presentation way………….Please post more on same way………:)Thanks

    Reply
    • ANKUSH PAWAR says

      June 20, 2016 at 11:25 AM

      Thank you so much…simple and useful explanation

      Reply
  10. Kalhan says

    March 16, 2015 at 4:58 PM

    Thank you so much…simple and useful explanation

    Reply
  11. Suresh says

    March 28, 2015 at 5:43 PM

    Nice Example

    Reply
  12. Arpitha says

    April 9, 2015 at 6:38 AM

    Good Explanation :)

    Thank you

    Reply
  13. surekha says

    July 15, 2015 at 9:48 AM

    When should use dynamic binding..Please explain clearly..

    Reply
  14. sam says

    July 16, 2015 at 10:32 AM

    HI,
    In the below
    Case 4:

    float mymethod(int a, float b)
    float mymethod(float var1, int var2)
    Result: Perfectly fine. Valid case for overloading. Sequence of the data types are different, first method is having (int, float) and second is having (float, int).
    It will get ambiguity.

    Reply
    • Aravind Reddy says

      December 2, 2015 at 4:12 AM

      Yes there is an ambiguity in case:4
      Because int can be implicitly converted into float.
      So the compiler is in a confused state to which method it should bind to

      Reply
  15. Ganesh NB says

    August 3, 2015 at 11:24 AM

    Great Explanation. Can u make it clear wheather java supports pass by reference.I know its no. still the value are updated in the calling function. can you explain how it occurs…?

    Thank you in advance for the answer

    Reply
  16. saad ismail says

    February 26, 2016 at 7:27 PM

    perfectly explain i,m loving it my all concepts are clear ….

    Reply
  17. Kemoji says

    March 1, 2016 at 6:08 AM

    Wonderful… Super cool stuff, made things easy…Thanks a Ton

    Reply
  18. Md Nehaluddin Haider says

    March 6, 2016 at 5:16 AM

    I tried reading many tutorials for OOPs and now I can say this is the best. Thank you so much buddy for this wonderful tutorial.

    Reply
  19. Ankit says

    April 13, 2016 at 9:57 AM

    Best Example everything is clear..
    Good Job

    Thanks

    Reply
  20. Jeremiah Nji says

    May 5, 2016 at 1:11 PM

    This was perfect! You explained it so clearly. Overloading is something I comprehend completely now!

    Thank you!

    Reply
  21. akshay says

    June 8, 2016 at 6:20 PM

    no one in our class taught me like this,
    Its simply awesome…. just now started my course,
    only by using your website only,
    i planned to go through daily,
    please help me for queries,

    Reply
  22. Luke says

    June 18, 2016 at 1:50 PM

    Very nice and uncomplicated way of explaining! Good job!!

    Reply
  23. Pankaj says

    June 20, 2016 at 2:52 PM

    your case 4 is not valid,

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

    when we call them with values, it will throw an error like error: reference to print is ambiguous h.print(10, 20);
    cause both are taking int value, that’s why reference variable is getting confused while selecting/calling them,

    Reply
  24. Kd says

    July 6, 2016 at 7:29 PM

    Excellent notes on method overloading! Kudos!

    Reply
  25. pooja says

    September 12, 2016 at 3:21 PM

    The concepts are clear and easy to understand. thank you

    Reply
  26. venkat says

    October 2, 2016 at 10:01 PM

    F:javaprograms>javac TestOverloading3.java
    F:javaprograms>java TestOverloading3
    Error: Could not find or load main class TestOverloading3
    its compiling successfully ,but its showing error like this while run.
    may i know why i am getting like this error

    Reply
  27. Shiva says

    November 13, 2016 at 7:07 PM

    Method Overloading Concepts are very clearly understood and got registered in my mind. Thanks a lot.

    Reply
  28. Deepanshu Jain says

    November 23, 2016 at 10:13 AM

    Just wanted to add one more case :
    If just return type is different !!

    public double myMethod(int num1, int num2)
    {
    System.out.println(“First myMethod of class Demo”);
    return num1+num2;
    }
    public float myMethod(int var1, int var2)
    {
    System.out.println(“Second myMethod of class Demo”);
    return var1-var2;
    }

    throw an error.

    Reply
    • Rakesh Ranjan says

      February 6, 2017 at 11:33 AM

      Already there in Question 2

      Reply
  29. Rajneesh Kaundal says

    January 12, 2017 at 1:04 PM

    Explained very well, Anyone can understand :)

    Reply
  30. EDISON MUALA says

    April 1, 2017 at 9:38 AM

    Thank you so much.the notes are clear and easy to follow.please keep up doing the good job, you have assisted me so much

    Reply
  31. sanad alteggaz says

    June 28, 2017 at 11:41 PM

    System.out.println(” thank you pro !!!”);

    Reply
  32. Himanshu Yadav says

    July 12, 2017 at 5:59 AM

    Can we take input at compile time in case of method overloading •••••?With the help of keyboard???

    Reply

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