Why is super.super.method(); not allowed in Java?

Question

I read this question and thought that would easily be solved (not that it isn't solvable without) if one could write:

@Override public String toString() {     return super.super.toString(); } 

I'm not sure if it is useful in many cases, but I wonder why it isn't and if something like this exists in other languages.

What do you guys think?

EDIT: To clarify: yes I know, that's impossible in Java and I don't really miss it. This is nothing I expected to work and was surprised getting a compiler error. I just had the idea and like to discuss it.

Answer 1

It violates encapsulation. You shouldn't be able to bypass the parent class's behaviour. It makes sense to sometimes be able to bypass your own class's behaviour (particularly from within the same method) but not your parent's. For example, suppose we have a base "collection of items", a subclass representing "a collection of red items" and a subclass of that representing "a collection of big red items". It makes sense to have:

public class Items {     public void add(Item item) { ... } }  public class RedItems extends Items {     @Override     public void add(Item item)     {         if (!item.isRed())         {             throw new NotRedItemException();         }         super.add(item);     } }  public class BigRedItems extends RedItems {     @Override     public void add(Item item)     {         if (!item.isBig())         {             throw new NotBigItemException();         }         super.add(item);     } } 

That's fine - RedItems can always be confident that the items it contains are all red. Now suppose we were able to call super.super.add():

public class NaughtyItems extends RedItems {     @Override     public void add(Item item)     {         // I don't care if it's red or not. Take that, RedItems!         super.super.add(item);     } } 

Now we could add whatever we like, and the invariant in RedItems is broken.

Does that make sense?

Answer 2

I think Jon Skeet has the correct answer. I'd just like to add that you can access shadowed variables from superclasses of superclasses by casting this:

interface I { int x = 0; } class T1 implements I { int x = 1; } class T2 extends T1 { int x = 2; } class T3 extends T2 {         int x = 3;         void test() {                 System.out.println("x=\t\t"          + x);                 System.out.println("super.x=\t\t"    + super.x);                 System.out.println("((T2)this).x=\t" + ((T2)this).x);                 System.out.println("((T1)this).x=\t" + ((T1)this).x);                 System.out.println("((I)this).x=\t"  + ((I)this).x);         } }  class Test {         public static void main(String[] args) {                 new T3().test();         } } 

which produces the output:

 x=              3 super.x=        2 ((T2)this).x=   2 ((T1)this).x=   1 ((I)this).x=    0 

(example from the JLS)

However, this doesn't work for method calls because method calls are determined based on the runtime type of the object.