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.

Solution 1

You can do:

super.alphaMethod1();

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

Solution 2

Simply use super.alphaMethod1();

See super keyword in java

Solution 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();
   }
}

Solution 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.

Solution 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)
    }
}

View more solutions

Comments

[Asif Mushtaq]

Some call super method at the end or some at the start why?

[Michał Šrajer]

In constructor, you need to call super() as a first statement (if you call it explicitly). On regular methods you call it wherever you want depending on your app logic needs. For example at the beginning if you want to add extra steps after call or at the end if you add extra check.

[Rudra]

@MichałŠrajer using super you cannot call method from class alpha into Test class which extends class beta

[U. Windl]

As a Java beginner I just wonder why super() does not have the magic to know in which method it is being used (Meaning: If foo() calls super(), it's obvious that super() means super.foo(). Constructors are just too special in Java IMHO) See Precursor in Eiffel, where it's done right.

[Pshemo]

Additional info: super.alphaMethod1(); can't be called from main method. This answer doesn't state it, but this call needs to be made from somewhere within non-static context of subclass: beta.

[abc123]

@Michal, this is a totally wrong answer. If you use super.alphaMethod1() you get a compiler error saying "cant invoke from static context, make main non-static". How is this answer upvoted??