Printing static variable without class name leads to an error

Question

public class ClearingDoubtsAboutStatic {

    static
    {
        System.out.println("Static Block1 Output: "+ClearingDoubtsAboutStatic.statVar);  //------Line 5
        statVar=20;  //-----Line 6
        System.out.println("Static Block1 Output: "+ClearingDoubtsAboutStatic.statVar); //------Line 7
        System.out.println("Static Block1 Output: "+statVar); //---Line 8

    }

    static int statVar=30;

    public static void main(String[] args) {

    }
}

What in my mind was that the line 7 and 8 will give the same output, but it is not the case.

My Question

what I don't understand is when we are able to initialize the static variable without the class name at line 6 but why we are not able to print it without the class name at line 8?

score 1 · Answer 1 · answered Dec 07 '18 at 09:46

The 2 rules you copy/paste are wrong, you should only consider:

Static declaration and static block initialization are considered in the order they appear in the source file

Thus, you can fix your issue, changing the order of the declaration, and your static initialization block:

static int statVar=30;

static
{
    System.out.println("Static Block1 Output: "+ statVar);  //------Line 5
    statVar=20;  //-----Line 6
    System.out.println("Static Block1 Output: "+ statVar); //------Line 7
    System.out.println("Static Block1 Output: "+statVar); //---Line 8

}

I totally got this, my question was totally different. I don't want my code to fix,I just wanted to understand why this is happening. I need an answer for the second last paragraph of my question — SanQA, Dec 07 '18 at 09:54

score 0 · Answer 2 · answered Dec 07 '18 at 10:10

0

I not quite sure about your case but in my IntelliJ when I try your code, I got this:

Error:(9, 55) java: illegal forward reference

answered Dec 07 '18 at 10:10

Hearen

7,420
4
53
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yeah, but I need the reason behind it. I really don't understand the logic "that you are able to modify that static variable without the class name but you aren't able to print it ". – SanQA Dec 07 '18 at 10:14

score 0 · Answer 3 · answered Dec 07 '18 at 10:22

0

Hmmm i'm not sure about that but i will share with you my guess. Initialization starts first for everything that is static in Class (in defined order). So your static{...} is will be initialized at first, then statVar will receive its value. First printing should show 0 as this is default int value. It works because your are referencing it by Class name and that is how static variables should be referenced. There is one static variable for Class, not for object. When you are trying to reference it without class name, your are treating it like just some field in your class and you shouldn't try to reference those before they are defined. So compiler is not allowing to do so.

answered Dec 07 '18 at 10:22

Mershel

542
1
9
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Then how you are able to modify the value of static variable without using class name at line 6? – SanQA Dec 07 '18 at 10:25
I guess that is because your are just giving it some starting value, initializing it. When you use it in `System.out.println` with `+` operator it has to be concatenated with your String so some casting needs to be done. – Mershel Dec 07 '18 at 10:32

Printing static variable without class name leads to an error

My Question

3 Answers3