CODEWITHSUNDEEP
java
Nigel Thomas
Nigel Thomas

Reputation: 1929

How does static variable in Java work?

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 ?

Upvotes: 2

Views: 167

Answers (6)

merlin2011
merlin2011

Reputation: 75629

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.

Upvotes: 1

user3218743
user3218743

Reputation: 589

 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.

Upvotes: 1

jbytecode
jbytecode

Reputation: 679

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.

Upvotes: 1

tsragravorogh
tsragravorogh

Reputation: 3173

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

Upvotes: 1

Mohsen Kamrani
Mohsen Kamrani

Reputation: 7457

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.

Upvotes: 1

Keppil
Keppil

Reputation: 46239

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);
}

Upvotes: 6

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