overriding - Can java call parent overridden method in other objects but not subtype?

Question

here is working java code

class Cup {
    public String sayColor() {
        return "i have a color .";
    }
}

class TCup extends Cup{
    public String sayColor(){
        System.out.println(super.getClass().getName());
        return super.sayColor()+"color is tee green.";
    }
}

class MyTCup extends TCup {
    public String sayColor(){
        System.out.println(super.getClass().getName());
        return super.sayColor()+"but brushed to red now!";
    }
}
class Test {
    public static void main(String[] args) {
        Cup c = new MyTCup();
        System.out.print(c.sayColor());
    }
}

and running the Test class prints

MyTCup
MyTCup
i have a color .color is tee green.but brushed to red now!

question 1: At the runtime, the type of object C is MyTCup, but it can always call the super method. Is there a method stack in the memory within MyTCup after initializing the object, and then can call through at runtime like the code ?

question 2: There is no way to call the super method in other objects. As I know ,c++ can cast to call parent method at any time. Why is it different in Java?

Answer 1

You can't call the super method in other objects - that would violate encapsulation. The whole point is that the object controls what its overridden methods do. For instance, you might override a collection's add method to throw an exception in certain circumstances, so it could ensure only "valid" items got added to the collection. That would be pointless if callers could just bypass it with a cast!

The only reason an object gets to call super.foo() for itself is to enable one call to be implemented by using the parent implementation. It's up to the code in the class to make sure it only ever does that sensibly. Again, to take the add-in-a-collection example, if the collection overrides add it would have to have some way of adding the validated item to the collection, which it would do with super.add().

Note that for the same reason of encapuslation, you can only call your parent implementation, not the grandparent implementation - so super.foo() is valid, but super.super.foo() isn't.