[java] how to inherit Constructor from super class to sub class

How to inherit the constructor from a super class to a sub class?

Superclass constructor CAN'T be inherited in extended class. Although it can be invoked in extended class constructor's with super() as the first statement.

You inherit class attributes, not class constructors .This is how it goes :

If no constructor is added in the super class, if no then the compiler adds a no argument contructor. This default constructor is invoked implicitly whenever a new instance of the sub class is created . Here the sub class may or may not have constructor, all is ok .

if a constructor is provided in the super class, the compiler will see if it is a no arg constructor or a constructor with parameters.

if no args, then the compiler will invoke it for any sub class instanciation . Here also the sub class may or may not have constructor, all is ok .

if 1 or more contructors in the parent class have parameters and no args constructor is absent, then the subclass has to have at least 1 constructor where an implicit call for the parent class construct is made via super (parent_contractor params) .

this way you are sure that the inherited class attributes are always instanciated .

Read about the super keyword (Scroll down the Subclass Constructors). If I understand your question, you probably want to call a superclass constructor?

It is worth noting that the Java compiler will automatically put in a no-arg constructor call to the superclass if you do not explicitly invoke a superclass constructor.

Say if you have

/**
 * 
 */
public KKSSocket(final KKSApp app, final String name) {
    this.app = app;
    this.name = name;
    ...
}

then a sub-class named KKSUDPSocket extending KKSSocket could have:

/**
 * @param app
 * @param path
 * @param remoteAddr
 */
public KKSUDPSocket(KKSApp app, String path, KKSAddress remoteAddr) {
    super(app, path, remoteAddr);
}

and

/**
 * @param app
 * @param path
 */
public KKSUDPSocket(KKSApp app, String path) {
    super(app, path);
}

You simply pass the arguments up the constructor chain, like method calls to super classes, but using super(...) which references the super-class constructor and passes in the given args.

Constructors are not inherited, you must create a new, identically prototyped constructor in the subclass that maps to its matching constructor in the superclass.

Here is an example of how this works:

class Foo {
    Foo(String str) { }
}

class Bar extends Foo {
    Bar(String str) {
        // Here I am explicitly calling the superclass 
        // constructor - since constructors are not inherited
        // you must chain them like this.
        super(str);
    }
}

Default constructors -- public constructors with out arguments (either declared or implied) -- are inherited by default. You can try the following code for an example of this:

public class CtorTest {
    public static void main(String[] args) {
        final Sub sub = new Sub();
        System.err.println("Finished.");
    }

    private static class Base {
        public Base() {
            System.err.println("In Base ctor");
        }
    }

    private static class Sub extends Base {
        public Sub() {
            System.err.println("In Sub ctor");
        }
    }
}

If you want to explicitly call a constructor from a super class, you need to do something like this:

public class Ctor2Test {
    public static void main(String[] args) {
        final Sub sub = new Sub();
        System.err.println("Finished.");
    }

    private static class Base {
        public Base() {
            System.err.println("In Base ctor");
        }

        public Base(final String toPrint) {
            System.err.println("In Base ctor.  To Print: " + toPrint);
        }
    }

    private static class Sub extends Base {
        public Sub() {
            super("Hello World!");
            System.err.println("In Sub ctor");
        }
    }
}

The only caveat is that the super() call must come as the first line of your constructor, else the compiler will get mad at you.