Interface question from JQPlus

[Jay Ashar]

interface I
{
     int i = 1, ii = Test.out("ii", 2);
}
 
interface J extends I
{
     int j = Test.out("j", 3), jj = Test.out("jj", 4);
}
 
interface K extends J
{
     int k = Test.out("k", 5);
}
 
class Test
{
     public static void main(String[] args) {  System.out.println(K.j);  }
     public static int out(String s, int i)
     {
          System.out.println(s + "=" + i);
          return i;
     }
}

What will be the output when class Test is run ?
Answer is
It will print j = 3, jj=4 and then 3.
Does anyone know why is it printing jj=4 too??
Edited by Corey McGlone: Added CODE Tags
[ March 08, 2004: Message edited by: Corey McGlone ]

[Gian Franco]

Hi Jay,
Check out the JLS chapter "12.4 Initialization of Classes and Interfaces".
I will do that myself as well. It explains your problem.
Best regards,
Gian Franco Casula
[ March 08, 2004: Message edited by: Gian Franco Casula ]

[Sekhar Kadiyala]

Hi Guys,

JLS chapter "12.4 Initialization of Classes and Interfaces"

Is it a book or online tutorial?

[Corey McGlone]

Originally posted by Sekhar Kadiyala:
Hi Guys,

Is it a book or online tutorial?

The JLS stands for Java Language Specification and is the ultimate source for all Java related questions (at least concerning the exam). You can find the entire specification online here.

[Sekhar Kadiyala]

Corey, Thank you.

[Jay Ashar]

I found the answer
"The reference to J.i is to a field that is a compile-time constant; therefore, it does not cause I to be initialized. The reference to K.j is a reference to a field actually declared in interface J that is not a compile-time constant; this causes initialization of the fields of interface J, but not those of its superinterface I, nor those of interface K. Despite the fact that the name K is used to refer to field j of interface J, interface K is not initialized."
Thanks Corey and Gian
[ March 08, 2004: Message edited by: Jay Patel ]

[Corey McGlone]

I'm glad you found it, Jay. I was hoping you'd take the initiative to hunt through the JLS for the answer - get used to using the JLS. The SCJP exam is based on it so, if you can find your way around it, you're much better off.