Java Static object/method issue

Question

I got a question about how to use static, I saw a sample:

public class Exe {
    static int i = 47;
    public void call() {
        for (i = 0; i < 3; i++) {
            if (i == 2) {
                System.out.println("\n");
            }
        }
    }

    public Exe() {
    }

    public static void main(String[] args) {
        Exe t1 = new Exe();
        Exe t2 = new Exe();
        t2.i = 60;
        System.out.println(t1.i);
        t1.call();
        System.out.println(t2.i);
        t2.call();
    }
}

When I tried to run it, it printed 60 3, I am wondering why t2.i here is 3, I do not know where the 3 comes from, also, the both results of t1.call() and t2.call() were not printed, please advise, thank you!

Man, you should format the code snippets properly, this one is not readable! — Adam Dyga
– Adam Dyga, Commented Sep 26, 2013 at 14:59
The results of t1.call() and t2.call() probably were printed. They would be blank lines. If you post the output, we could see. — Reinstate Monica -- notmaynard
– Reinstate Monica -- notmaynard, Commented Sep 26, 2013 at 15:02
If the static keyword is giving you issues, try this. Start a new project, take out the static keyword, but leave everything else the same, then run it. Compre the output from one to the other and see if that helps clear the issue up. — nhgrif
– nhgrif, Commented Sep 26, 2013 at 15:09

Adi Inbar · Accepted Answer · 2013-10-04 19:16:11Z

2

for(i=0; i<3;i++){
    if(i==2){
        System.out.println("\n");
    }
}

Your static variable Assigned/incremented here.

Not the i (which you are assuming, it's different) in for loop.

To clear the clouds, Just take another variable called j and do the looping.

for (int j = 0; j < 3; j++) {
    if (j == 2) {
        System.out.println("\n");
    }
}

edited Oct 4, 2013 at 19:16

Adi Inbar

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answered Sep 26, 2013 at 14:59

Suresh Atta

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2 Comments

Peter Lawrey Over a year ago

+1 This is where using a debugger would have answered the question

chrylis -cautiouslyoptimistic- Over a year ago

There isn't any i "in the for loop"; there's only the one i anywhere.

nhgrif · Accepted Answer · 2013-09-26 15:05:04Z

1

static means that every instance of the class has access to the same, single instance of the variable.

When you create t1, t1.i is initialized to 47. Then you create t2. t2.i and t1.i are the same variable, so whether you do t1.i = 60 or t2.i = 60, they're BOTH equal to 60.

So before you do t1.call(); or t2.call(), the first thing you do is print out t1.i, which is 60, as per the line t1.i = 60;.

Then you run t1.call() which runs through the for loop. The for loop exits when i can't pass the test i < 3, and since i is an integer, this happens as soon as i is incremented to 3.

After you've run t1.call(), i is now equal to 3. This means both t1.i and t2.i since static means there's only one copy of i across all instances of the Exe class here. So you print out t2.i and it is equal to 3, as it should be.

Hope this helps.

answered Sep 26, 2013 at 15:05

nhgrif

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2 Comments

Wang Pei The Dancer Over a year ago

Does it mean that the call of t1.call() change the value of i?

nhgrif Over a year ago

Yes. Because for the for loop in the call() method of your Exe class uses the variable i, the loop re-initializes i to 0. And the static keyword mean that t1.i and t2.i refer to the same variable.

chrylis -cautiouslyoptimistic- · Accepted Answer · 2013-09-26 15:00:15Z

1

The 3 comes from your for loop, which is reusing the same static int i that you're manually setting to 60 before you call call(). The results of the println probably are being printed, but they're just blank lines.

answered Sep 26, 2013 at 15:00

chrylis -cautiouslyoptimistic-

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Comments

Aniket Thakur · Accepted Answer · 2013-09-26 15:03:08Z

0

for (i = 0; i < 3; i++)

In this statement instead of treating i as a local variable it is taken to be class variable which you defined. As for class variables only one copy is maintained for all the instance after the for loop finished(i=3) value of you class variable i remains to be 3.

answered Sep 26, 2013 at 15:03

Aniket Thakur

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Comments

RyanS · Accepted Answer · 2013-09-26 15:05:11Z

0

Your static variable is the same across all objects, this is an effect of static. This means that when you call t2.i = 60; both t1.i and t2.i are set equal to 60. Then when you call t1.call(); you again change both objects. As others have explained your for loop is setting the variable i = 3.

It's because it's a static variable that the assignment affects both objects

answered Sep 26, 2013 at 15:05

RyanS

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Comments

Nagendra Varma · Accepted Answer · 2013-09-26 16:30:53Z

Static members doesn't reflect the object state!

All objects of a class will have the same copy(mean to say same value) of the static variables. If one object changes the value of a static variable, since the static variable holds the same value across all the objects, the static variable value will be updated across all the objects of that class. This rule reflects your result.

See the below example:

public class StaticTest {

    static int staticA = 10;

    int intA;

    public StaticTest(int intA){
        this.intA = intA;
    }

    public static void main(String[] args) {
        StaticTest test1 = new StaticTest(10);
        System.out.println(test1.intA);
        System.out.println(test1.staticA);
        StaticTest test2 = new StaticTest(20);
        System.out.println(test2.intA);
        System.out.println(test2.staticA);
    }
}

Below is the outpput:

10 10 20 10

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