CODEWITHSUNDEEP
java
Priyank B
Priyank B

Reputation: 33

method overriding with wrapper class

public class A
{
    public void display(int i)
    {
        System.out.println("Inside A");
    }
}

public class B extends A
{
    public void display(Integer i)
    {
        System.out.println("Inside B");
    }
}


public class starter
{
    public static void main (String args[])
    {
        A a = new B();
        a.display(5);
        System.out.println("So now you know or not");
    }
}

Output : Inside A

Can somebody explain this output? Normally child method should be called. How does Java behave here when we have a wrapper class and a primitive class using inheritance?

Upvotes: 1

Views: 1594

Answers (1)

Mena
Mena

Reputation: 48404

B#display does not override A#display, as the signature is different.

The fact ints can be boxed into Integers is not relevant here.

You could easily verify this by using the @Override annotation.

Since the reference type of a is A, the method is resolved with the exact match for a literal integer (your given 5 argument), which is int, therefore A#display is invoked.

You can still force the invocation of B#display by using this idiom (not for production code):

((B)a).display(new Integer(5));

This casts your a variable as a B type, hence allowing visibility of B's display method within context.

It also passes an Integer rather than an int, thus employing the signature of B's display method and allowing resolution to that method.

Upvotes: 4

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