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.

stackoverflow.com/questions/40480/…

Romain Hippeau
– Romain Hippeau

2015-01-08 15:43:21 +00:00
Commented Jan 8, 2015 at 15:43 — Romain Hippeau
– Romain Hippeau, Commented Jan 8, 2015 at 15:43

Sotirios Delimanolis · Accepted Answer · 2015-01-08 15:39:25Z

2

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

answered Jan 8, 2015 at 15:39

Sotirios Delimanolis

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4 Comments

jarosik Over a year ago

so how to access an 'a' instance variable from modify()? You can't use 'this' in static context right?

Sotirios Delimanolis Over a year ago

@jarosik You can qualify it. Main.a.

jarosik Over a year ago

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.

Sotirios Delimanolis Over a year ago

@jarosik The concept is known as scope.

Thusitha Thilina Dayaratne · Accepted Answer · 2015-01-08 15:43:23Z

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 8, 2015 at 15:43

Thusitha Thilina Dayaratne

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Comments

Balázs Édes · Accepted Answer · 2015-01-08 15:46:11Z

0

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.

answered Jan 8, 2015 at 15:46

Balázs Édes

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