overriding protected method of Superclass

Question

In the below example why does the String b prints null and String c prints "gg".

Correct me if I am wrong, whenever a subclass (BClass) overrides a protected method (i.e initClass()) of the superclass (AClass).If you instantiate the subclass. The superclass must make use of overriden method specified by the subclass.

public class Example {

    public class AClass {

        private String a;

        public AClass() {
            initClass();
        }

        protected void initClass() {
            a = "randomtext";
        }
    }

    public class BClass extends AClass {

        private String b = null; 
        private String c;          


        @Override
        protected void initClass() {
            b = "omg!";
            c = "gg";
        }

        public void bValue() {
            System.out.println(b);   // prints null
            System.out.println(c);  // prints "gg"
        }
    }

    public static void main(String[] args) {
        Example.BClass b = new Example().new BClass();
        b.bValue();

    }

}

Answer 1

As of the JSF 12.5

In the example you can see the execution order. The first steps are the callings of the Constructor down to the Object constructor. Afterwards this happens:

Next, all initializers for the instance variables of class [...] are executed.

Since your instance variable b is initialized to null it will be null again afterwards

Answer 2

This is happening because the superclass constructor is called before the fields of ClassB is initialized. Hence the initClass() method is called which sets b = "omg!" but then again when the super class constructor returns, b is initialized to the value declared in ClassB which is null.

To debug, put a break point and go step by step, you will find that b is first set to null and then changes to omg! and then comes back to null.