Java Inheritance - calling superclass method

Question

Lets suppose I have the following two classes

public class alpha {

    public alpha(){
        //some logic
    }

    public void alphaMethod1(){
        //some logic
    }
}

public class beta extends alpha {

    public beta(){
        //some logic
    }

    public void alphaMethod1(){
        //some logic
    }
}

public class Test extends beta
{
     public static void main(String[] args)
      {
        beta obj = new beta();
        obj.alphaMethod1();// Here I want to call the method from class alpha.
       }
}

If I initiate a new object of type beta, how can I execute the alphamethod1 logic found in class alpha rather than beta? Can I just use super().alphaMethod1() <- I wonder if this is possible.

Autotype in Eclipse IDE is giving me the option to select alphamethod1 either from class alpha or class beta.

Answer 1

You can do:

super.alphaMethod1();

Note, that super is a reference to the parent class, but super() is its constructor.

Answer 2

Simply use super.alphaMethod1();

See super keyword in java

Answer 3

You can't call alpha's alphaMethod1() by using beta's object But you have two solutions:

solution 1: call alpha's alphaMethod1() from beta's alphaMethod1()

class Beta extends Alpha
{
  public void alphaMethod1()
  {
    super.alphaMethod1();
  }
}

or from any other method of Beta like:

class Beta extends Alpha
{
  public void foo()
  {
     super.alphaMethod1();
  }
}

class Test extends Beta 
{
   public static void main(String[] args)
   {
      Beta beta = new Beta();
      beta.foo();
   }
}

solution 2: create alpha's object and call alpha's alphaMethod1()

class Test extends Beta
{
   public static void main(String[] args)
   {
      Alpha alpha = new Alpha();
      alpha.alphaMethod1();
   }
}

Answer 4

It is possible to use super to call the method from mother class, but this would mean you probably have a design problem. Maybe B.alphaMethod1() shouldn't override A's method and be called B.betaMethod1().

If it depends on the situation, you can put some code logic like :

public void alphaMethod1(){
    if (something) {
        super.alphaMethod1();
        return;
    }
    // Rest of the code for other situations
}

Like this it will only call A's method when needed and will remain invisible for the class user.

Answer 5

Solution is at the end of this answer, but before you read it you should also read what is before it.

---

What you are trying to do would break security by allowing skipping possible validation mechanisms added in overridden methods.

For now lets imagine we can invoke version of method from superclass via syntax like

referenceVariable.super.methodName(arguments)

If we have classes like

class ItemBox{ //can sore all kind of Items
    public void put(Item item){
        //(1) code responsible for organizing items in box 
    }

    //.. rest of code, like container for Items, etc.
}

class RedItemsBox extends ItemBox {//to store only RED items

    @Override
    public void put(Item item){ //store only RED items
        if (item.getColor()==Color.RED){
            //(2) code responsible for organizing items in box
        }
    }
}

As you see RedItemsBox should only store RED items.

Regardless which of the below we use

ItemBox     box = new RedItemsBox();
RedItemsBox box = new RedItemsBox();

calling

box.put(new BlueItem());

will invoke put method from RedItemsBox (because of polymorphism). So it will correctly prevent BlueItem object from being placed in RedItemBox.

But what would happen if we could use syntax like box.super.put(new BlueItem())?

Here (assuming it would be legal) we would execute version of put method from ItemBox class.
BUT that version doesn't have step responsible for validating Item color. This means that we could put any Item into a RedItemBox.

Existence of such syntax would mean that validation steps added in subclasses could be ignored at any time, making them pointless.

---

There IS a case where executing code of "original" method would make sense.

And that palce is inside overriding method.

Notice that comments //(1) .. and //(2).. from put method of ItemBox and RedItemBox are quite similar. Actually they represent same action...

So it makes sense to reuse code from "original" method inside overriding method. And that is possible via super.methodName(arguments) call (like from inside put of RedItemBox):

@Override
public void put(Item item){ //store only RED items
    if (item.getColor()==Color.RED){
        super.put(item); // <<<--- invoking code of `put` method 
                         //        from ItemBox (supertype)
    }
}

Answer 6

Whenever you create child class object then that object has all the features of parent class. Here Super() is the facilty for accession parent.

If you write super() at that time parents's default constructor is called. same if you write super.

this keyword refers the current object same as super key word facilty for accessing parents.

Answer 7

beta obj = new beta(); Since you have created beta object , you cant refer directly to alphamethod1 of alpha object. It can be modified as

class Beta extends Alpha
{
  public void alphaMethod1()
  {
    super.alphaMethod1();
  }
}