Variable Initialization...

[Shah Chunky]

Hello all...
Can someone explain me how the output of below program is Generated ?
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(J.i);
System.out.println(K.j);
}
static int out(String s, int i) {
System.out.println(s + "=" + i);
return i;
}
}

produces the output:

1
j=3
jj=4
3
Thanks.

[Cindy Glass]

I thought the JLS described it quite well:

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.

System.out.println(J.i); // The i variable is inherited from I but since it is a constant - you don't have to go through the whole process of initializing I, just reference it and get on with things.
System.out.println(K.j); // now this is more complicated. Since the variable holds something other than a constant, the first thing that we have to do is initialize J so that we can figure out what to DO to come up with the value of j.
Initializing interface J causes variable j to be initialized first: executing Test.out("j",3) which prints j=3, and setting the variable j to 3 (note the return type on the method). Now jj is initialized executing Test.out("jj",4) which prints jj=4 and sets the variable jj to 4.
Now that initialization is done we can evaluate and execute the println for K.j which prints the current value of j which is 3.

[Vinod Mehta]

(1) System.out.println(J.i);
will print output 1
(2) System.out.println(J.i); will print
j=3
jj=4
3