method overriding

[maggie karve]

   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("super"));
		
		subFoo = new SubCafe4Java();
		superFoo = subFoo;
		System.out.println (superFoo.get("super"));
	}
} 

why is superclass method called???

[Wouter Oet]

Because you're not overriding the method get and you're using a reference of type SuperCafe4Java.

Don't forget to Quote Your Sources

[Deepika Joshi]

this is method overloading.

[Deepika Joshi]

    
   class SuperCafe4Java {  
    public Object get (Object o) {  
        return ("SuperCafe4Java");  
    }  
}  

class SubCafe4Java extends SuperCafe4Java {  
    @Override
    public Object get (String o) {  
        return ("SubCafe4Java");  
    }  
}  

try @Override, it confirms if it is overriding or not.

[Minhaj Mehmood]

maggie joseph wrote:

   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("super"));
		
		subFoo = new SubCafe4Java();
		superFoo = subFoo;
		System.out.println (superFoo.get("super"));
	}
} 

why is superclass method called???

because you are not overriding the method, you are overloading it.
remember the rule of overriding the number of parameter(s) and the type of parameter(s) must be same where the return type can be co-variant of an overridden method.