[Exercise] Prove that base class constructor is always called

[Pho Tek]

Fellow java'ers,
TIJ has a question that reads:
Prove that the base-class constructor are
(a) always called, and
(b) called before derived-class constructors.
My idea for (a) follows. Your suggestions welcomed.
Assume that the base class is A and the derived class is B.
e.g. class B extends A{ ... }
I could make the constructor private in the base class.
That way I know for sure that the instantiation of class B may fail if it can't find a ctor in A.

/// B.java
class A{
private A() { System.out.print("A ctor"); }
}
public class B extends A{}
Compiling this gives me the error:

No Constructor matching A() found in class A.

Is this formal enough ? Have I covered all the bases?
Anyone want to give (b) a try?
Thanks
Pho

[Manfred Leonhardt]

Hi Pho,
I don't think your example answers any of the questions. Getting a compiler error rarely proves anything! The following code can be used to prove both a and b.

class A
{
  public A()
  {
    System.out.println( "Super class constructor" );
  }
}
public class B extends A
{
  /*
   * Having constructors with prints will show super before sub
   * Remove both constructors to prove always called (compiler default).
   */
  public B()
  {
    this( 10 );
  }
  public B( int i )
  {
    System.out.println( "Sub class constructor" );
  }
  public static void main( String[] arg )
  {
    B b = new B();
  }
}

Regards,
Manfred.