Strange code on Java Static Initialization Blocks

Question

When going through the JLS 8.3.2.3 I wasn't able to understand the following code.

class Z {
static { i = j + 2; }
static int i, j;
static { j = 4; }
}

The code is resulting in the error Cannot reference a field before it is defined

But if I change the code to

class Z {
static { i = 2; }
static int i, j;
static { j = 4; }
}

The code is getting compiled. But in both the cases the variable definition is after the initialization block. What is the mystery behind this?

Answer 1

You can assign to a value earlier than its declaration - you just can't read it. So this fails too:

static { System.out.println(j + 2); }
static int j;

Whereas this is fine:

static { j = 5; }
static int j;

One of the four conditions in section 8.3.2.3 for an invalid usage is:

• The usage is not on the left hand side of an assignment.

(The double-negatives in that section are making my head hurt, but I think it relevant!)

To be honest, that part of the spec is one of the worst I've seen - it's really unclear. But the upshot is that you can assign but not read :)

Answer 2

Actually,Its an compiler basics,

the assignment statements are executed from right to left.

e.g.i=2;

it means 2 is assigned to i,and 2 is constant hence no need to declare it

In the other hand if we write

i=j+2;

it will compile j first and then assign it to the i hence it causes the error because j is not defined yet.

Answer 3

In i = j + 2; you use the variable j that is initialize in a different static block.

You should put everything togheter, in only one static block, the j initialization and the expression to get the code compiled. This code works:

public class Z {

    static int i, j;
    static { j = 4; 
            i = j + 2; }

}

Davide