working of static in following code

Question

class Test {
    static int x = 11;
    private int y = 33;
    public void method(int x) {
        Test t = new Test();
        this.x = 22;
        y = 44;
        System.out.println(Test.x);
        System.out.println(t.x);
        System.out.println(t.y);
        System.out.println(y);
    }
    public static void main(String args[]) {
        Test t = new Test();
        t.method(5);
    }
}

Here the code is ok and it gives output like: 22 22 33 44

but I don 't understand:

Why Test.x isn't 11 and give output like: 11 22 33 44
Why t.y and y isn't a 44 and 44 and gives output like:11 22 44 44
Why t.method(5) is not executed in this code?

Steve Smith · Answer 1 · 2017-09-12T13:35:46.147

0

1) x is static, there is only one of them. So Test.x and this.x are the same var.

2) y is not static, so they are separate vars.

3) method() is executed. Otherwise you would get no output.

BTW, I think you might be confusing yourself by having two instances of Test, both called t. I would rename one.

edited Sep 12 '17 at 13:35

answered Sep 12 '17 at 13:29

Steve Smith

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score 0 · Answer 2 · answered Sep 12 '17 at 13:29

The difference between a static field and an instance field is that the static field is not instance-dependent.

Look here:

static int x = 11;
private int y = 33;

The static field x will be the same for every instance of Test while the field of y will be instance specific. That's why you see these results.

score 0 · Answer 3 · answered Sep 12 '17 at 14:02

First, you need to understand what static keyword does, here's a good question to start with: What does the 'static' keyword do in a class?

Now that you know that static makes a variable a member of the Class and not of the instance, you will know that if an instance of class modify a static field of that class, another instance will see the modified value, we can go through your questions:

Why Test.x isn't 11 and give output like: 11 22 33 44

Test.x refers to this variable: static int x = 11;, which you're modifying its value to 22 here:

this.x = 22;

In this case, you have a local variable, which you're receiving as a parameter int x. When using this keyword, you're accessing the instance of this class and as it's a static variable, there's no instance but you access it.

So, when you call this.x = 22 is like saying Test.x = 22.

Why t.y and y isn't a 44 and 44 and gives output like: 11 22 44 44

It's due to y not being static, so it has it's own value. When you call new Test();, the variable t creates a new instance having a variable called y initialized to 33, because of this:

private int y = 33;

So, when you call t.y it prints the above result because it's a different instance.

When you print y, then you're calling the variable y which is in use in the program right now (and it's a different instance from the one above), and you changed its value to 44 before printing it.

That's why the first this.y prints 33 and y prints 44

Why t.method(5) is not executed in this code?

It's indeed being executed, or you wouldn't see any output at all, but you don't get a 5 because you're never using the local variable in the parameters int x. You could try writing

System.out.println(x);

So you get that value.

working of static in following code

3 Answers3