显式转换运算符和多态

Question

I have a small issue with polymorphism and explicit casting. Here is the problem: I have 3 public classes:

  package x;

public class Shape
{
 public void draw(){
     System.out.println("draw a Shape");
 }
}

package x;

public class Circle1 extends Shape
{

}

package x;

public class Circle2 extends Circle1
{
    @Override
    public void draw ()
    {
       System.out.println("draw Circle2");
    }
}

And my Demo class is:

package x;

public class Demo
{
    public static void main (String[] args)
    {

        Shape s1=new Circle2();
        s1.draw();
        ((Shape)s1).draw();

    }
}

The output of this code is:

draw Circle2
draw Circle2

I understand the polymorphic behavior of s1.draw() that invokes the draw() method on Circle2 . The last line confuses me, when I do explicit casting on s1 like so: ((Shape)s1) , it means that the expression ((Shape)s1) is a reference of type Shape because of the explicit casting,right? And if that is so, why then the code ((Shape)s1).draw(); invokes the draw() method in Circle2 and not the one in Shape?

Thanks :)

Answer 1

Casting does not change what the object is , it simply changes how the compiler should look at that object.

So even if you "think" of a Circle to be "just a shape" it still stays a Circle.

The only way for a Circle1 to call Shape 's draw would be from within Circle1 by a call to super.draw()

Answer 2

When you explicitly (or implicitly) cast an object to a super type the methods will always obey the implementation of the class used to construct the object. In your case, because your object was constructed with new Circle2() , it will always be a Circle2, even if you cast it. In fact, there is an easy way for you to see that. Just add the following lines to your code:

Shape s2 = (Shape) s1;
System.out.println(s2.getClass().getName());

This prints the exact class that your s2 object (casted from s1) belongs to :) you will see it is still a Circle2.

If you want to call the super type implementation of the draw() method you need to call super.draw() instead.