Passing variable into methods (Java)

Question

I'm new to Java, and I have a problem. I've copied some code off a tutorial for Android, and now I want to pass an integer variable into the method run(), so I can increment it for each loop and then catch it outside the background Thread. How would I do that?

int gg= 0;    
Thread background = new Thread(new Runnable() {
                    public void run() {
                        try {

                            while (pBarDialog.getProgress() <= 100) {

                                Thread.sleep(100);
                                gg++; // the increment here
                                progressHandler.sendMessage(progressHandler
                                        .obtainMessage());


                            }
                            if (pBarDialog.getProgress() == 100) {
                                pBarDialog.dismiss();

                            }

                        } catch (java.lang.InterruptedException e) {
                            // if something fails do something smart
                        }
                    }

                });
          //catch gg here

What do you mean by 'catching' the variable? Btw, I believe that pBarDialog.dismiss() should be called on the UI thread, namely in Handler. — Ash
– Ash, Commented Jan 13, 2012 at 8:00
anything you want to pass to an inner class can be done by using "final" (but then you must copy the value of the variable to another one if you want to modify it) — njzk2
– njzk2, Commented Jan 13, 2012 at 9:14

KV Prajapati · Accepted Answer · 2012-01-13 08:20:56Z

3

You can't specify argument to the run() method. You may declare int variable as field and use it in inner classes.

public class TestActivity extends Activity
{
   private volatile int no;
   .....

}

EDIT: (Suggestion from @alf) You can use volatile modifier with field so changed value can be seen immediately by all other threads.

edited Jan 13, 2012 at 8:20

answered Jan 13, 2012 at 7:59

KV Prajapati

94.8k20 gold badges151 silver badges188 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

alf Over a year ago

Note it should better be volatile to ensure visibility.

JB Nizet Over a year ago

He MUST use volatile, else the counter value makes no sense.

stacker · Accepted Answer · 2012-01-13 08:58:18Z

1

Have your own class and pass the counter using its constructor, I haven't tried that, but I would start with something like that:

class MyThread implements Runnable {

   private volatile int counter;

   public MyThread( int counter ) {
       this.counter = counter;
   }

   public void run() {
   ...
   }

   public getCounter() {
      return counter;
   }
}

MyThread mt = new MyThread( 10 );
Thread t = new Thread( mt );
t.start();

// after some time
t.getCounter();

edited Jan 13, 2012 at 8:58

answered Jan 13, 2012 at 8:04

stacker

69.1k29 gold badges143 silver badges214 bronze badges

2 Comments

JB Nizet Over a year ago

Why use Integer rather than int? The counter should be volatile, or it should always be accessed from a synchronized block. Else, the main thread could aways see 10 as the value, even if the background thread sets it to 1000000.

stacker Over a year ago

@JBNizet Thanks I fixed it, I was on my way to work as I recognized my failure. Fortunatly I wasn't that heavily downvoted as I thought.

score 0 · Accepted Answer · 2012-01-13 08:20:41Z

0

private volatile int gg;

public void myMethod() {
    Thread background = new Thread(new Runnable() {

        @Override
        public void run() {
            try {
                while (pBarDialog.getProgress() <= 100) {
                    Thread.sleep(100);
                    gg++; // the increment here
                    progressHandler.sendMessage(progressHandler.obtainMessage());
                }
                if (pBarDialog.getProgress() == 100) {
                    pBarDialog.dismiss();
                }
            } catch (java.lang.InterruptedException e) {
                // if something fails do something smart
            }
        }

    });

    System.out.println(gg);
}

edited Jan 13, 2012 at 8:20

answered Jan 13, 2012 at 8:00

user882347

6 Comments

alf Over a year ago

You cannot ++ a final variable, can you?

user882347 Over a year ago

I've never had a problem with it. I've edited my answer to include an alternative.

JB Nizet Over a year ago

Calling this.gg = gg won't work, since gg is final. Why do you make it final since the goal is to change its value?

JB Nizet Over a year ago

You can't ++ a final argument either.

JB Nizet Over a year ago

Now it's almost correct. gg must be volatile, or the main thread might always see the initial value. And using public fields is bad practice.

|

alf · Accepted Answer · 2012-01-13 08:37:13Z

0

If I were you, I'd be looking into AtomicInteger, namely the incrementAndGet() method.

Making gg a field will indeed give the thread access to gg, and volatile will make the changes visible, but since your intentions are not clear, I cannot be sure that you don't have other threads incrementing the same value: you don't have atomicity, so as soon as you have more than one thread doing gg++, you're likely to get wrong results.

answered Jan 13, 2012 at 8:37

alf

8,5811 gold badge26 silver badges45 bronze badges

Collectives™ on Stack Overflow

Passing variable into methods (Java)

4 Answers 4

2 Comments

2 Comments

6 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

2 Comments

2 Comments

6 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related