Java classes and static blocks

Question

class Hello12 {
    static int b = 10;
    static {
        b = 100;
    }
}

class sample {
    public static void main(String args[]) {
        System.out.println(Hello12.b);
    }
}

On running above code the output comes as 100 because when I called Hello class, static block is executed first setting the value of b to 100 and displaying it. But when i write this code:

class Hello12 {
    static {
         b = 100;
    }
    static int b = 10;
}

class sample {
    public static void main(String args[]) {
        System.out.println(Hello12.b);
    }
}

Here the output comes as 10. I am expecting answer as 100 because once the static block is executed it gave b the value as 100. so when in main(), I called Hello.b it should have referred to b (=100). How is the memory allocated to b in both the codes?

score 50 · Accepted Answer · answered Jul 22 '14 at 18:58

50

In the "Detailed Initialization Procedure" for classes, Section 12.4.2 of the JLS states:

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.

This means that it's as if the first example was:

b = 10;
b = 100;

And the second example was:

b = 100;
b = 10;

The last assignment "wins", explaining your output.

answered Jul 22 '14 at 18:58

rgettman

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@kocko That's okay. Let's help get rgettman the [Populist](http://stackoverflow.com/help/badges/62/populist) badge. Unfortunately, this means the accepted answer needs to have 10 upvotes... – Justin Jul 22 '14 at 22:43
5

Why does it compile if it's run in 'textual order'? Surely `b` hasn't been declared when it is set in the static block? – Holloway Jul 23 '14 at 09:16
Because the compiler *like always* splits Variable DECLARATION from assignment. Like in the good old days, every Variable is first declared, a nice place in Memory is found and made cozy, before any assignments happen ;-) – Falco Jul 23 '14 at 15:52

score 16 · Answer 2 · edited Jul 23 '14 at 02:46

16

Static blocks and static variables are initialized in the order in which they appear in the source. If your code is:

class Hello12 {

  static int b = 10;
  static {
     b = 100;
  }

}

The result is 100.

edited Jul 23 '14 at 02:46

doppelgreener

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answered Jul 22 '14 at 18:59

betteroutthanin

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score 5 · Answer 3 · answered Jul 22 '14 at 19:18

Besides answering the question of how the code is executed in what order, I am guessing you also want to know why a static block can refer to a static variable that has not been textually declared/executed yet.

While section 12.4.2 of the JLS does explain that static blocks and static variable are executed in the textual order that they appear, section 8.3.3 of the JLS explains when you can reference what, and you can see that the condition of The use is not on the left hand side of an assignment; fails, allowing your static block in the second example to refer to a static variable that has not textually in order been declared/executed yet.

score 4 · Answer 4 · answered Jul 22 '14 at 18:59

4

Statics are evaluated in the order they appear in the program.

answered Jul 22 '14 at 18:59

Bathsheba

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Java classes and static blocks

4 Answers4

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