How does interface inheritance work ??

Question

I have the following code. The printout is
Main
w
x
w
I don't understand how this is working. Clearly interface inheritance is working differently than normal class inheritance. Please explain.
thanks in advance.
kris.
The code follows
interface C
     char w = DD.echo ('w');
     char x = DD.echo ('x');
interface D extends C
     char y = DD.echo('y');
     char z = DD.echo('z');
     char a = DD.echo(w);
class DD implements D
     static char echo(char c)
          System.out.print(c + "\n");
          return c;
     public static void main (String args[])
          System.out.println("Main");
          DD dd = new DD();
          System.out.println(w + "\n");
}

Answer 1

The interface only needs loaded if one of its members
is accessed (although this is probably implemented
differently on different JVMs).No.
The interface is loaded.
It is not initialized. There is a difference. See section 12 of the JLS.
Assuming the above is true, it appears the following
is happening
1. C is loaded because w is accessed from DD�s main .
This loading causes �w� to be echoed and then �x� to
be echoed since loading the interface initializes the
members which in turn call the echo method (I am
assuming that this only happends when the members are
not �constants�).
2. The main method then echoes the w variable (which
is �w�).
Having said this, I would think this code would behave
differently on different JVMs so I am not sure how it
could even be a �good� theoretical question (unless
this is documented in the JLS).It is in the JLS.
Also, loading the the interface may not really be what
is happening; the initialization code may be in-lined
in the class (this could be tested by printing w twice
in main � really strange behaviour).No. It can't inline the initilization of one class in another class.