Java: Calling a super method which calls an overridden method

I think of it this way

+----------------+
|     super      |
+----------------+ <-----------------+
| +------------+ |                   |
| |    this    | | <-+               |
| +------------+ |   |               |
| | @method1() | |   |               |
| | @method2() | |   |               |
| +------------+ |   |               |
|    method4()   |   |               |
|    method5()   |   |               |
+----------------+   |               |
    We instantiate that class, not that one!

Let me move that subclass a little to the left to reveal what's beneath... (Man, I do love ASCII graphics)

We are here
        |
       /  +----------------+
      |   |     super      |
      v   +----------------+
+------------+             |
|    this    |             |
+------------+             |
| @method1() | method1()   |
| @method2() | method2()   |
+------------+ method3()   |
          |    method4()   |
          |    method5()   |
          +----------------+

Then we call the method
over here...
      |               +----------------+
 _____/               |     super      |
/                     +----------------+
|   +------------+    |    bar()       |
|   |    this    |    |    foo()       |
|   +------------+    |    method0()   |
+-> | @method1() |--->|    method1()   | <------------------------------+
    | @method2() | ^  |    method2()   |                                |
    +------------+ |  |    method3()   |                                |
                   |  |    method4()   |                                |
                   |  |    method5()   |                                |
                   |  +----------------+                                |
                   \______________________________________              |
                                                          \             |
                                                          |             |
...which calls super, thus calling the super's method1() here, so that that
method (the overidden one) is executed instead[of the overriding one].

Keep in mind that, in the inheritance hierarchy, since the instantiated
class is the sub one, for methods called via super.something() everything
is the same except for one thing (two, actually): "this" means "the only
this we have" (a pointer to the class we have instantiated, the
subclass), even when java syntax allows us to omit "this" (most of the
time); "super", though, is polymorphism-aware and always refers to the
superclass of the class (instantiated or not) that we're actually
executing code from ("this" is about objects [and can't be used in a
static context], super is about classes).

In other words, quoting from the Java Language Specification:

The form super.Identifier refers to the field named Identifier of the current object, but with the current object viewed as an instance of the superclass of the current class.

The form T.super.Identifier refers to the field named Identifier of the lexically enclosing instance corresponding to T, but with that instance viewed as an instance of the superclass of T.

In layman's terms, this is basically an object (*the** object; the very same object you can move around in variables), the instance of the instantiated class, a plain variable in the data domain; super is like a pointer to a borrowed block of code that you want to be executed, more like a mere function call, and it's relative to the class where it is called.

Therefore if you use super from the superclass you get code from the superduper class [the grandparent] executed), while if you use this (or if it's used implicitly) from a superclass it keeps pointing to the subclass (because nobody has changed it - and nobody could).

The keyword super doesn't "stick". Every method call is handled individually, so even if you got to SuperClass.method1() by calling super, that doesn't influence any other method call that you might make in the future.

That means there is no direct way to call SuperClass.method2() from SuperClass.method1() without going though SubClass.method2() unless you're working with an actual instance of SuperClass.

You can't even achieve the desired effect using Reflection (see the documentation of java.lang.reflect.Method.invoke(Object, Object...)).

[EDIT] There still seems to be some confusion. Let me try a different explanation.

When you invoke foo(), you actually invoke this.foo(). Java simply lets you omit the this. In the example in the question, the type of this is SubClass.

So when Java executes the code in SuperClass.method1(), it eventually arrives at this.method2();

Using super doesn't change the instance pointed to by this. So the call goes to SubClass.method2() since this is of type SubClass.

Maybe it's easier to understand when you imagine that Java passes this as a hidden first parameter:

public class SuperClass
{
    public void method1(SuperClass this)
    {
        System.out.println("superclass method1");
        this.method2(this); // <--- this == mSubClass
    }

    public void method2(SuperClass this)
    {
        System.out.println("superclass method2");
    }

}

public class SubClass extends SuperClass
{
    @Override
    public void method1(SubClass this)
    {
        System.out.println("subclass method1");
        super.method1(this);
    }

    @Override
    public void method2(SubClass this)
    {
        System.out.println("subclass method2");
    }
}



public class Demo 
{
    public static void main(String[] args) 
    {
        SubClass mSubClass = new SubClass();
        mSubClass.method1(mSubClass);
    }
}

If you follow the call stack, you can see that this never changes, it's always the instance created in main().

You can only access overridden methods in the overriding methods (or in other methods of the overriding class).

So: either don't override method2() or call super.method2() inside the overridden version.