Why call super() in a constructor?

Question

I'm dealing with a class which extends JFrame .

It's not my code and it makes a call to super before it begins constructing the GUI. I'm wondering why this is done since I've always just accessed the methods of the superclass without having to call super();

Answer 1

There is an implicit call to super() with no arguments for all classes that have a parent - which is every user defined class in Java - so calling it explicitly is usually not required. However, you may use the call to super() with arguments if the parent's constructor takes parameters, and you wish to specify them. Moreover, if the parent's constructor takes parameters, and it has no default parameter-less constructor, you will need to call super() with argument(s).

An example, where the explicit call to super() gives you some extra control over the title of the frame:

class MyFrame extends JFrame
{
    public MyFrame() {
        super("My Window Title");
        ...
    }
}

Answer 2

A call to your parent class's empty constructor super() is done automatically when you don't do it yourself. That's the reason you've never had to do it in your code. It was done for you.

When your superclass doesn't have a no-arg constructor, the compiler will require you to call super with the appropriate arguments. The compiler will make sure that you instantiate the class correctly. So this is not something you have to worry about too much.

Whether you call super() in your constructor or not, it doesn't affect your ability to call the methods of your parent class.

As a side note, some say that it's generally best to make that call manually for reasons of clarity.

Answer 3

We can access super class elements by using super keyword

Consider we have two classes, Parent class and Child class, with different implementations of method foo. Now in child class if we want to call the method foo of parent class, we can do so by super.foo(); we can also access parent elements by super keyword.

    class parent {
    String str="I am parent";
    //method of parent Class
    public void foo() {
        System.out.println("Hello World " + str);
    }
}

class child extends parent {
    String str="I am child";
    // different foo implementation in child Class
    public void foo() {
        System.out.println("Hello World "+str);
    }

    // calling the foo method of parent class
    public void parentClassFoo(){
        super.foo();
    }

    // changing the value of str in parent class and calling the foo method of parent class
    public void parentClassFooStr(){
        super.str="parent string changed";
        super.foo();
    }
}


public class Main{
        public static void main(String args[]) {
            child obj = new child();
            obj.foo();
            obj.parentClassFoo();
            obj.parentClassFooStr();
        }
    }

Answer 4

它只是调用超类的默认构造函数。

Answer 5

We use super keyword to call the members of the Superclass.

As a subclass inherits all the members (fields, methods, nested classes) from its parent and since Constructors are NOT members (They don't belong to objects. They are responsible for creating objects), they are NOT inherited by subclasses.

So we have to explicitly give the call for parent constructor so that the chain of constructor remains connected if we need to create an object for the superclass. At the time of object creation, only one constructor can be called. Through super, we can call the other constructor from within the current constructor when needed.

If you are thinking why it's there for a class that is not extending any other class, then just remember every class follows object class by default. So it's a good practice to keep super in your constructor.

Note: Even if you don't have super() in your first statement, the compiler will add it for you!

Answer 6

We can Access SuperClass members using super keyword

If your method overrides one of its superclass's methods, you can invoke the overridden method through the use of the keyword super . You can also use super to refer to a hidden field (although hiding fields is discouraged). Consider this class, Superclass:

public class Superclass {

    public void printMethod() {
        System.out.println("Printed in Superclass.");
    }
}

// Here is a subclass, called Subclass, that overrides printMethod() :

public class Subclass extends Superclass {

    // overrides printMethod in Superclass
    public void printMethod() {
        super.printMethod();
        System.out.println("Printed in Subclass");
    }
    public static void main(String[] args) {
        Subclass s = new Subclass();
        s.printMethod();    
    }
}

Within Subclass, the simple name printMethod() refers to the one declared in Subclass, which overrides the one in Superclass. So, to refer to printMethod() inherited from Superclass, Subclass must use a qualified name, using super as shown. Compiling and executing Subclass prints the following:

Printed in Superclass.
Printed in Subclass

Answer 7

None of the above answers answer the 'why'. Found a good explanation here :

A subclass can have its own private data members, so a subclass can also have its own constructors.

The constructors of the subclass can initialize only the instance variables of the subclass. Thus, when a subclass object is instantiated the subclass object must also automatically execute one of the constructors of the superclass.

Cheers

Answer 8

as constructor is not a part of class, so while calling it cannot be implemented, by using SUPER() we can call the members and memberfunctions in constructor.