Interface variables

[karimkhan pathan]

  public class TestKK {
	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;
	}
}
interface I {
	int i = 10;
	int ii = TestKK.out("ii", 2);
}
interface J extends I {
	int j = TestKK.out("j", 3);
	int jj = TestKK.out("jj", 4);
	
}
interface K extends J {
	int k = TestKK.out("k", 5);
}
  

output :
10
j=3
jj=4
3

Can you tell me why jj=4 is printing when i call K.j

[Bupjae Lee]

I think this example is from this part of Java Language Specification

I recommand you visit and read this link.



In short: K.j actually is J.j and trying to use J.j invokes iniitialization of interface J.

During initialization, both J.j and J.jj should be initialized, so both j=3 and jj=4 are printed.

[karimkhan pathan]

Then why is not printing when we invoke J.i(First SOP)

[Sagar Rohankar]

karimkhan pathan wrote:Then why is not printing when we invoke J.i(First SOP)

JLS wrote:The reference to J.i is to a field that is a compile-time constant; therefore, it does not cause I to be initialized.

Quote from the same link.

[Seetharaman Venkatasamy]

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.

enough information . Study twice,if you not understand .

[Sagar Rohankar]

But, isn't that tricky question which need the "thorough" knowledge of JLS.

[Bupjae Lee]

In my opinion, Java "compile-time constant" invokes some corner case.

Moreover, I think referring JLS gives the most clear answer.