Overloaded method

[Anthony Karta]

why lines #1 and #2 are calling SuperCafe4Java get() method?
the actual object is SubCafe4Java and it has more specific get(String) method.

    
class SuperCafe4Java {
public Object get (Object o) {
return ("SuperCafe4Java");
}
}
 
class SubCafe4Java extends SuperCafe4Java {
public Object get (String o) {
return ("SubCafe4Java");
}
}
 
class TestCafe4Java {
public static void main (String[] arguments) {
SuperCafe4Java superFoo;
SubCafe4Java subFoo;
 
superFoo = new SubCafe4Java();
System.out.println (superFoo.get(new String("super")));//#!
 
subFoo = new SubCafe4Java();
        System.out.println (subFoo.get("sub"));
superFoo = subFoo;
System.out.println (superFoo.get("super")); //#2
}
}

[Manju Devarla]

In this program get() method is overloaded so


SuperCafe4Java superFoo = new SubCafe4Java();

System.out.println (superFoo.get(new String("super")));//#!

#1 : superFoo.get() method is called based on the reference not on runtime object type.

Output of this program is

SuperCafe4Java
SubCafe4Java
SuperCafe4Java

[NDP Prasad]

If you change the object to String you will get the desired output.

class SuperCafe4Java {public Object get ([B]String[/B] o) {return ("SuperCafe4Java");}}

[Anuragk kushwaha]

Dear NDP..

waht you did ? , just overriding.

with overriding we can achieve runtime polymorphism while using overloading we can achieve compile time polymorphism.

it means in case of overloading method called using refeneces.


---please correct me if i am wrong.