Lecture 5

Lecture # 5

Method Overloading

If a class has multiple methods having same name but different in parameters, it is known as Method Overloading. If we have to perform only one operation, having same name of the methods increases the readability of the program.

Suppose you have to perform addition of the given numbers but there can be any number of arguments, if you write the method such as a(int,int) for two parameters, and b(int,int,int) for three parameters then it may be difficult for you as well as other programmers to understand the behavior of the method because its name differs.

So, we perform method overloading to figure out the program quickly.

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)

  1. Data type of parameters.
    For example:

add(int, int)

add(int, float)

  1. 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)

 

Q) Why Method Overloading is not possible by changing the return type of method only?

In java, method overloading is not possible by changing the return type of the method only because of ambiguity. Let’s see how ambiguity may occur:

  1. classAdder{
  2. staticint add(int a,int b){return a+b;}
  3. staticdouble add(int a,int b){return a+b;}
  4. }
  5. classTestOverloading3{
  6. publicstatic void main(String[] args){
  7. out.println(Adder.add(11,11));//ambiguity
  8. }}

Test it Now

Output:

Compile Time Error: method add(int,int) is already defined in class Adder

 

EXTRA study material

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:

aa 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:

a5

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 dispI’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 → longshort → int → longint → long → float → doublefloat → doublelong → 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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