static instance variable pass by value

Question

I thought static primitive variables in java should work differently than non-static when passed to a method:

public class Main {

    private static int a;

    public static void main(String[] args) {
        modify(a);
        System.out.println(a);
    }

    static void modify(int a){ 
        a++;
    }
}

The output is 0 which is understandable for primitives passed by value, but why static doesn't mean anything here? I would expect 1 as an output.

Maybe a silly question but I am confused.

http://stackoverflow.com/questions/40480/is-java-pass-by-reference-or-pass-by-value — Romain Hippeau, Jan 08 '15 at 15:43

score 2 · Answer 1 · answered Jan 08 '15 at 15:39

2

The name a in your modify method is referring to the local method parameter, not the static variable.

answered Jan 08 '15 at 15:39

Sotirios Delimanolis

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so how to access an 'a' instance variable from modify()? You can't use 'this' in static context right? – jarosik Jan 08 '15 at 15:43
@jarosik You can qualify it. `Main.a`. – Sotirios Delimanolis Jan 08 '15 at 15:44
Other valid solution I came up with: you can just rename the method argument to: modify (int b) or anything else than 'a' and the plain 'a++' would work. – jarosik Jan 10 '15 at 23:46
@jarosik The concept is known as _scope_. – Sotirios Delimanolis Jan 11 '15 at 00:27

score 0 · Answer 2 · answered Jan 08 '15 at 15:43

0

You are having a shadowing variable in your static method when a++ get executed it will increase the value of that method local variable by 1

static variable default value is 0 and will not get affect.

if you want that to be changed use

Main.a++;

answered Jan 08 '15 at 15:43

Thusitha Thilina Dayaratne

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score 0 · Answer 3 · answered Jan 08 '15 at 15:46

If you really want to, you could solve this by an int wrapper, like AtomicInteger:

public class Main {

    private static final AtomicInteger a = new AtomicInteger(0);

    public static void main(String[] args) {
        modify(a);
        System.out.println(a);
    }

    static void modify(AtomicInteger  a){ 
        a.getAndIncrement(); // "eqvivalent" of a++
    }
}

Your current code, takes an int, and because how Java works, it recieves a copy of your static a, and has no effect on your static field.

static instance variable pass by value

3 Answers3