包装类重写方法

Question

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?

Answer 1

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

The fact int s can be boxed into Integer s 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.