Method Overloading in Java
- Different ways
to overload the method
- By changing the
no. of arguments
- By changing the
datatype
- Why method
overloading is not possible by changing the return type
- Can we overload
the main method
- method
overloading with Type Promotion
If a
class have multiple methods by same name but different 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 behaviour of the method because its
name differs. So, we perform method overloading to figure out the program
quickly.
Advantage of method
overloading?
Method overloading increases the readability of the
program.
Different ways to overload the method
|
There
are two ways to overload the method in java
|
|
1. By
changing number of arguments
2. By
changing the data type
|
In java,
Methood Overloading is not possible by changing the return type of the method.
1)Example
of Method Overloading by changing the no. of arguments
In this
example, we have created two overloaded methods, first sum method performs
addition of two numbers and second sum method performs addition of three
numbers.
1.
class Calculation{
2. void sum(int a,int b){System.out.println(a+b);}
3. void sum(int a,int b,int c){System.out.println(a+b+c);}
4.
5. public static void main(String args[]){
6. Calculation obj=new Calculation();
7. obj.sum(10,10,10);
8. obj.sum(20,20);
9.
10. }
11. }
Output:30
40
2)Example
of Method Overloading by changing data type of argument
In this
example, we have created two overloaded methods that differs in data type. The
first sum method receives two integer arguments and second sum method receives
two double arguments.
1.
class Calculation2{
2. void sum(int a,int b){System.out.println(a+b);}
3. void sum(double a,double b){System.out.println(a+b);}
4.
5. public static void main(String args[]){
6. Calculation2 obj=new Calculation2();
7. obj.sum(10.5,10.5);
8. obj.sum(20,20);
9.
10. }
11. }
Output:21.0
40
Que) Why Method Overloaing is not possible by
changing the return type of method?
In
java, method overloading is not possible by changing the return type of the
method because there may occur ambiguity. Let's see how ambiguity may occur:
because
there was problem:
1.
class Calculation3{
2. int sum(int a,int b){System.out.println(a+b);}
3. double sum(int a,int b){System.out.println(a+b);}
4.
5. public static void main(String args[]){
6. Calculation3 obj=new Calculation3();
7. int result=obj.sum(20,20); //Compile Time Error
8.
9. }
10. }
int result=obj.sum(20,20);
//Here how can java determine which sum() method should be called
Can we overload main() method?
Yes, by
method overloading. You can have any number of main methods in a class by
method overloading. Let's see the simple example:
1.
class Overloading1{
2. public static void main(int a){
3. System.out.println(a);
4. }
5.
6. public static void main(String args[]){
7. System.out.println("main() method invoked");
8. main(10);
9. }
10. }
Output:main() method invoked
10
Method
Overloading and TypePromotion
One
type is promoted to another implicitly if no matching datatype is found. Let's
understand the concept by the figure given below:
As displayed in the above
diagram, byte can be promoted to short, int, long, float or double. The short
datatype can be promoted to int,long,float or double. The char datatype can be
promoted to int,long,float or double and so on.
Example of Method Overloading with
TypePromotion
1.
class OverloadingCalculation1{
2. void sum(int a,long b){System.out.println(a+b);}
3. void sum(int a,int b,int c){System.out.println(a+b+c);}
4.
5. public static void main(String args[]){
6. OverloadingCalculation1 obj=new OverloadingCalculation1();
7. obj.sum(20,20);//now second int literal will be promoted to long
8. obj.sum(20,20,20);
9.
10. }
11. }
Output:40
60
Example of Method Overloading with
TypePromotion if matching found
If
there are matching type arguments in the method, type promotion is not
performed.
1.
class OverloadingCalculation2{
2. void sum(int a,int b){System.out.println("int arg method invoked");}
3. void sum(long a,long b){System.out.println("long arg method invoked");}
4.
5. public static void main(String args[]){
6. OverloadingCalculation2 obj=new OverloadingCalculation2();
7. obj.sum(20,20);//now int arg sum() method gets invoked
8. }
9. }
Output:int arg method invoked
Example of Method Overloading with
TypePromotion in case ambiguity
If
there are no matching type arguments in the method, and each method promotes
similar number of arguments, there will be ambiguity.
1.
class OverloadingCalculation3{
2. void sum(int a,long b){System.out.println("a method invoked");}
3. void sum(long a,int b){System.out.println("b method invoked");}
4.
5. public static void main(String args[]){
6. OverloadingCalculation3 obj=new OverloadingCalculation3();
7. obj.sum(20,20);//now ambiguity
8. }
9. }
Output:Compile Time Error
One
type is not de-promoted implicitly for example double cannot be depromoted to
any type implicitly.
No comments:
Post a Comment