overloading + inheritance + widening

[Akila Sathyan]

Hi,
came across this code in a mock exam.
----------------

public class Circle extends Shape
{
   public void scale(int x)
    {
        System.out.println("Scale Circle int ");
    }
    
    public static void main(String args[])
    {
        Shape s = new Circle();
        s.scale(1);
    }
 
}

class Shape
{
    public void scale(long x)
    {
        System.out.println("Scale Shape long");
    }
 
}

-----------
Since the parameter to the method "scale()" was '1' - an int literal, I assumed that the method in the Circle class will be invoked. But the method in the base class Shape gets invoked. Can someone please explain?

[edit]Add code tags. CR[/edit]
[ October 04, 2008: Message edited by: Campbell Ritchie ]

[Henry Wong]

came across this code in a mock exam.

Also... Please Quote Your Sources.

Thanks,
Henry

[Henry Wong]

Since the parameter to the method "scale()" was '1' - an int literal, I assumed that the method in the Circle class will be invoked. But the method in the base class Shape gets invoked. Can someone please explain?

The reference type is of Shape. The method (signature of) that it will call -- including which overloaded method, is determined at compile time, using the reference type. Based on the reference, there is only one scale() method, which takes a long, hence, that is the method that is called.

Henry
[ October 04, 2008: Message edited by: Henry Wong ]

[Akila Sathyan]

Sorry about the protocol goof-ups. I've updated my profile (I did have to give up a little bit of my culture - I'm from I-only-got-one-name culture!)

Here is the mock exam link:
http://www.irixtech.com/category/java/scjp-quiz