Order of execution lines of code

Question

I don't get this :

In a ShakeListener class, I execute a routine in the containing class.

The routine is :

    public void showWord(){
        myShakeListener.stop();     

        flipper.showNext();

        v.vibrate(countdown5, -1);

        try {
            Thread.sleep(5000);
        } catch (InterruptedException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        myShakeListener.start();
    }

Strange thing (to me, still a beginner), is that the thread sleeps BEFORE the next view is shown. Why is that?

What I want to accomplish : user shakes phone -> viewflipper flips to next -> Phone is unresponsive to shaking for 5 seconds -> user shakes phone -> viewflipper flips to next...

thnx

score 1 · Accepted Answer · answered Jul 18 '12 at 20:26

1

The problem is that the viewflipper is probably another thread. You're hitting a race condition. Better option is to spawn a thread for 5 seconds that sets a boolean called something like "noshake" to true when it starts and sets it false it when it's done. Check if noshake == false before allowing another shake.

Does that makes sense?

answered Jul 18 '12 at 20:26

Rick Mangi

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Yep! I get it.. will implement shortly and mark this answer correct after that. Thnx – Kees Koenen Jul 19 '12 at 10:20

score 1 · Answer 2 · answered Jul 18 '12 at 20:28

It's because your code is blocking the UI thread. You should do something like this:

Handler mHandler = new Handler();
public void showWord(){
    myShakeListener.stop();     

    flipper.showNext();

    v.vibrate(countdown5, -1);

    mHandler.postAtTime(new Runnable() {
        @Override
        public void run() {
            myShakeListener.start();
        }
    }, 5000);
}

Order of execution lines of code

2 Answers2