How does static variable in Java work?

Question

When the below code is executed

class Foo {

    static int size = 7;

    static void changeIt(int size) {
        size = size + 200;
        System.out.println("size in changeIt is " + size);
    }

    public static void main(String[] args) {

        System.out.println("size = " + size);
        changeIt(size);
        System.out.println("size after changeIt is " + size);
    }
}

the answer will be

size = 7
size in changeIt is 207
size after changeIt is 7

The answer which I expected was

size  after changeIt is 207

Why the values of static variable is differing in the changeIt method and the main method ?

This has nothing to do with static variables. The size argument in changeIt is "hiding" the static field. — Mattias Buelens
– Mattias Buelens, Commented Apr 13, 2014 at 18:02

Keppil · Accepted Answer · 2014-04-13 18:02:35Z

6

Why the values of static variable is differing in the changeIt method and the main method ?

The values are different because you didn't change the static variable size, you changed the local variable size. Change the code to:

static void changeIt() {
    size = size + 200;
    System.out.println("size in changeIt is " + size);
}

Or, if you want to use an argument with the same name, you can specify that you are using the static class variable by qualifying it with the class name:

static void changeIt(int size) {
    Foo.size = size;
    System.out.println("size in changeIt is " + Foo.size);
}

answered Apr 13, 2014 at 18:02

Keppil

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Comments

Mohsen Kamrani · Accepted Answer · 2014-04-13 18:02:52Z

1

change size = size + 200 to

Foo.size = size + 200;

size is local in your method, then you change the variable that you want.

The problem is that the parameter name size is equal to name of the static field size so when you point it in that method compiler considers the local variable.

answered Apr 13, 2014 at 18:02

Mohsen Kamrani

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1 Comment

geoand Over a year ago

Would probably be better as Foo.size

tsragravorogh · Accepted Answer · 2014-04-13 18:06:14Z

1

Your method:

static void changeIt(int size) 
{
    size = size + 200;
    System.out.println("size in changeIt is " + size);
}

Does not change your internal static member variable. It takes an integer and adds 200 to it. If you just want this method to add 200 to your static member variable your function should look like this:

static void changeIt() 
{
    size = size + 200;
    System.out.println("size in changeIt is " + size);
}

To learn more about static variables, look here: http://en.wikipedia.org/wiki/Static_variable

answered Apr 13, 2014 at 18:06

tsragravorogh

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Comments

jbytecode · Accepted Answer · 2014-04-13 18:06:15Z

1

Because you use the parameter name 'size' which is same as the static variable. When you type

size = size + 1;

you change the value of local variable. Use

ClassName.size += 1;

instead.

answered Apr 13, 2014 at 18:06

jbytecode

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Comments

user3218743 · Accepted Answer · 2014-04-13 18:08:50Z

1

 static void changeIt(int size) {
 size = size + 200;
 System.out.println("size in changeIt is " + size);
  }

Assume that you replace "size" with "x". Then you can simply understand the answer you get is correct. Just like that "size" in the changeIt() method does not even compare with the static "size". Read bit more stuff about scope of variables.

answered Apr 13, 2014 at 18:08

user3218743

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Comments

merlin2011 · Accepted Answer · 2014-04-13 18:12:19Z

1

Java and many other languages do what is called local variable shadowing. That means the local variable size will "cover up" the class-level variable size.

You can work around this by either using this.size, or removing the size parameter from your method signature.

If you want more details on Java specifically, you can consult the JLS 6.4.1.

edited Apr 13, 2014 at 18:12

answered Apr 13, 2014 at 18:03

merlin2011

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