Java static final field initialization order

Question

I tried to understand the behavior of initialization order when static fields are initialized with a reference to the same enclosing class object.

public class Test {

        static final Test t=new Test();
        static int a=5;

        Test(){
            System.out.println("a="+a);
        }

        public static void main(String[] args) {
            new Test();
        }

    }

Output of above piece of code is:

a=0
a=5

If I modify variable a to anything else other than plain static:

static final a=5;
a=5;
final a=5;

The output is:

a=5
a=5

Why is this behavior?

Note that the output is a=5 & a=5 even when both t & a are declared as static final in which case t precedes the declaration of a

Answer 1

static final members are initialized before other static members.

non final static members are initialized in order of appearance

Therefore, in your first case :

    static Test t=new Test();
    static int a=5;

The constructor is first called before a is initialized, so a=0 is displayed.

In the second case, static final a is initialized before t, so a=5 is displayed when the first instance of Test is created. When a is not static, it is initialized prior to the execution of the constructor, so again a=5 is displayed.

Regarding the edit in your question.

Looking at section 12.4.2 of the JLS :

6. Then, initialize the final class variables and fields of interfaces whose values are compile-time constant expressions (§8.3.2.1, §9.3.1, §13.4.9, §15.28).

...

9. Next, execute either the class variable initializers and static initializers of the class, or the field initializers of the interface, in textual order, as though they were a single block.

You see that final class variables (i.e. static final) are initialized before the rest of the static variables only if their values are compile time constant expressions. 5 is a constant expression. new Test() is not. Therefore a is initialized before t even if both are static final.