Why call super() in a constructor?

问题

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();

回答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");
        ...
    }
}

回答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.

回答3:

It simply calls the default constructor of the superclass.

回答4:

We can access super class elements by using super method

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() method.

    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();
        }
    }

回答5:

We can Access SuperClass members using super()

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

来源：https://stackoverflow.com/questions/10508107/why-call-super-in-a-constructor

标签

java

superclass