Karatas, FERHAT
I would love to change the world, but they won't give me the source code!
Accessing Superclass Members
12.2.2007 11:33:48 - Filed under : Java
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 {

        public void printMethod() { //overrides printMethod
            System.out.println("Printed in Subclass");
        public static void main(String[] args) {
        Subclass s = new Subclass();
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
Subclass Constructors
The following example illustrates how to use the super keyword to invoke a superclass's constructor. Here is the MountainBike (subclass) constructor that calls the superclass constructor and then adds initialization code of its own:
public MountainBike(int startHeight, int startCadence, int startSpeed, int startGear) {
            super(startCadence, startSpeed, startGear);
            seatHeight = startHeight;
Invocation of a superclass constructor must be the first line in the subclass constructor.
The syntax for calling a superclass constructor is
super(parameter list);
With super(), the superclass no-argument constructor is called. With super(parameter list), the superclass constructor with a matching parameter list is called.

Note: If a constructor does not explicitly invoke a superclass constructor, the Java compiler automatically inserts a call to the no-argument constructor of the superclass. If the super class does not have a no-argument constructor, you will get a compile-time error. Object does have such a constructor, so if Object is the only superclass, there is no problem.

If a subclass constructor invokes a constructor of its superclass, either explicitly or implicitly, you might think that there will be a whole chain of constructors called, all the way back to the constructor of Object. In fact, this is the case. It is called constructor chaining, and you need to be aware of it when there is a long line of class descent.

Related Terminology : "Base" keyword in C#
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