Java: Calling a super method which calls an overridden method

Question

public class SuperClass
{
    public void method1()
    {
        System.out.println(\"superclass method1\");
        this.method2();
    }

    public void method2(         


        

Answer 1

Since the only way to avoid a method to get overriden is to use the keyword super, I've thought to move up the method2() from SuperClass to another new Base class and then call it from SuperClass:

class Base 
{
    public void method2()
    {
        System.out.println("superclass method2");
    }
}

class SuperClass extends Base
{
    public void method1()
    {
        System.out.println("superclass method1");
        super.method2();
    }
}

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

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

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

Output:

subclass method1
superclass method1
superclass method2

Answer 2

Further more extended the output of the raised question, this will give more insight on the access specifier and override behavior.

            package overridefunction;
            public class SuperClass 
                {
                public void method1()
                {
                    System.out.println("superclass method1");
                    this.method2();
                    this.method3();
                    this.method4();
                    this.method5();
                }
                public void method2()
                {
                    System.out.println("superclass method2");
                }
                private void method3()
                {
                    System.out.println("superclass method3");
                }
                protected void method4()
                {
                    System.out.println("superclass method4");
                }
                void method5()
                {
                    System.out.println("superclass method5");
                }
            }

            package overridefunction;
            public class SubClass extends SuperClass
            {
                @Override
                public void method1()
                {
                    System.out.println("subclass method1");
                    super.method1();
                }
                @Override
                public void method2()
                {
                    System.out.println("subclass method2");
                }
                // @Override
                private void method3()
                {
                    System.out.println("subclass method3");
                }
                @Override
                protected void method4()
                {
                    System.out.println("subclass method4");
                }
                @Override
                void method5()
                {
                    System.out.println("subclass method5");
                }
            }

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

            subclass method1
            superclass method1
            subclass method2
            superclass method3
            subclass method4
            subclass method5

Answer 3

class SuperClass
{
    public void method1()
    {
        System.out.println("superclass method1");
        SuperClass se=new SuperClass();
        se.method2();
    }

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


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

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

calling

SubClass mSubClass = new SubClass();
mSubClass.method1();

outputs

subclass method1
superclass method1
superclass method2

Answer 4

During my research for a similar case, I have been ending up by checking the stack trace in the subclass method to find out from where the call is coming from. There are probably smarter ways to do so, but it works out for me and it's a dynamic approach.

public void method2(){
        Exception ex=new Exception();
        StackTraceElement[] ste=ex.getStackTrace();
        if(ste[1].getClassName().equals(this.getClass().getSuperclass().getName())){
            super.method2();
        }
        else{
            //subclass method2 code
        }
}

I think the question to have a solution for the case is reasonable. There are of course ways to solve the issue with different method names or even different parameter types, like already mentioned in the thread, but in my case I dindn't like to confuse by different method names.

Answer 5

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).

Answer 6

To summarize, this points to current object and the method invocation in java is polymorphic by nature. So, method selection for execution, totally depends upon object pointed by this. Therefore, invoking method method2() from parent class invokes method2() of child class, as the this points to object of child class. The definition of this doesn't changes, irrespective of whichever class it's used.

PS. unlike methods, member variables of class are not polymorphic.