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.
Very clear. Excellent examples. Thanks a lot
Thanks.. very good and easy to understand
Excellent Example keep it up and thanks :-)
very clear….
The examples helped very much. The examples were clear, and to the point. Great explanation. :D
Very well presentation. Thanks a lot.
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 ..
that’s right
I think what is written in the article is correct. Don’t understand why Shipra agrees.
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.
There couldn’t be a more detailed and useful explanation of method overloading than this.
Thank you!
You have described in well manner i like that your presentation way………….Please post more on same way………:)Thanks
Thank you so much…simple and useful explanation
Thank you so much…simple and useful explanation
Nice Example
Good Explanation :)
Thank you
When should use dynamic binding..Please explain clearly..
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.
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
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
perfectly explain i,m loving it my all concepts are clear ….
Wonderful… Super cool stuff, made things easy…Thanks a Ton
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.
Best Example everything is clear..
Good Job
Thanks
This was perfect! You explained it so clearly. Overloading is something I comprehend completely now!
Thank you!
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,
Very nice and uncomplicated way of explaining! Good job!!
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,
Excellent notes on method overloading! Kudos!
The concepts are clear and easy to understand. thank you
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
Method Overloading Concepts are very clearly understood and got registered in my mind. Thanks a lot.
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.
Already there in Question 2
Explained very well, Anyone can understand :)
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
System.out.println(” thank you pro !!!”);
Can we take input at compile time in case of method overloading •••••?With the help of keyboard???