Question

In the following example:

class Base {    
    int  x=10;  

    Base() {    
      show();
    }  

    void show() {   
        System.out.print ("Base Show " +x + "  ");
    }  
}  

class Child extends Base {   
    int x=20;  

    Child() {
        show();
    }  

    void show() {    
        System.out.print("Child Show " + x +"  ") ; 
    }  

    public static void main( String s[ ] ) {   
        Base obj = new Child();   
    }  
} 

• Why is the output as shown below

Child Show 0  Child Show 20

• I thought constructors can only access instance members once its super constructors have completed.

I think what is happening here is that the super constructor is calling the child’s show() method because this method was overridden in Child. as it has been overridden but why is the value of x 0 and why is it able to access this method before the super constructor has completed?

Answer 1

I think what is happening here is that the super constructor is calling the child’s show() method because this method was overriden in Child.

That is correct

but why is the value of x 0

because it’s not initialized yet (x of Child)

and why is it able to access this method before the super constructor has completed?

That’s exactly why in a constructor you should never call a method, which can be overridden (non-final public and protected).

Edit:

The strange thing here is that everything has default/ package-private visibility. This can have some strange effects. See: http://www.cooljeff.co.uk/2009/05/03/the-subtleties-of-overriding-package-private-methods/

I recommend to avoid overriding methods with default visibility if possible (you can prevent this by declaring them final).