Simple use of static in java

Question

I understand the purpose of static keyword but regretfully I can't figure out this simple code. Don't know why? I'm expecting the answer to be really simple.

public class VariableScope {

    int x=y;
    static int y=5;

    public static void main(String[] args) {
        System.out.println(new VariableScope().x); 
    }
}

How x gets printed as 5 when y was assigned to it at an earlier stage?

What do you expect it to print? When you do think the static assignment happens? Please enumerate explicitly what you expect to happen? — Boris the Spider
– Boris the Spider, Commented Nov 9, 2014 at 18:15
Because all statics are initialised before everything else (in your case, local x) when the programme is launched. — ha9u63a7
– ha9u63a7, Commented Nov 9, 2014 at 18:16
@hagubear this is an utterly fallacious over-simplification. — Boris the Spider
– Boris the Spider, Commented Nov 9, 2014 at 18:17
This is example 8.3.2.2-1 in the JLS Initializers for Instance Variables The adjacent sections are also good reads: Restrictions on the use of Fields during Initialization and Initializers for Class Variables. — user289086
– user289086, Commented Nov 9, 2014 at 18:17

Costi Ciudatu · Accepted Answer · 2014-11-09 18:17:24Z

9

The static initialization static int y = 5 happens during class loading.

The x = y assignment happens during the instance construction new VariableScope(); so at this point y has the value 5 already.

answered Nov 9, 2014 at 18:17

Costi Ciudatu

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1 Comment

Boris the Spider Over a year ago

Simple and correct! If only there were more answers like this!

Michal Krasny · Accepted Answer · 2014-11-09 18:22:29Z

2

It's because order the lines of the code does not matter here. Static member is initialized first, then x is initialized.

Here is detailed specification of the initialization order: Java Language Specification 12.4.2. Detailed Initialization Procedure

edited Nov 9, 2014 at 18:22

answered Nov 9, 2014 at 18:19

Michal Krasny

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