Threads

[Lovleen Gupta]

Here is the code:

   class TestThread extends Thread{
    public void restart(){
        startMe();
    }
    public void startMe(){
        synchronized(this){
        notifyAll();
        System.out.println("Trying to notify");
        }
    }
 
public void run(){
    try{
       synchronized(this){
       wait();
       System.out.println("Notified");
        }
    } catch(InterruptedException e){}
}
    public static void main(String[] args){
    TestThread t1 = new TestThread();
    t1.start();
    t1.restart();
    
    }
} 

The output, I think, should be "Trying to Notify" and "Notified" with no certainity of the order in which they will appear.

However, the output is only "Trying to Notify".
Please explain why so?

Thanks.

[Chandra Bhatt]

Hi Guptajee,

Change your main with this code: An tell what you got!

 
public static void main(String[] args){
    TestThread t1 = new TestThread();
    
    t1.start();
        
    try{
    Thread.sleep(1000);
    }catch(InterruptedException ie) {
    ie.printStackTrace();
    System.exit(1);
    }
    
    t1.restart();
}

waiting...
synchronized("Guptajee") {
try {
wait("Guptajee");
}catch(InterruptedException ie) {
ie.printlnStacktrace();
System.exit(1);
}
}
Regards,
cmbhatt
[ April 09, 2007: Message edited by: Chandra Bhatt ]

[Henry Wong]

However, the output is only "Trying to Notify".
Please explain why so?

Calling the start() method, will create a new thread that will eventually call the run() method. But there is no guarantee that it will call the run() method right away.

In your case, if the main thread returns from the start() method, and calls the restart() method, before the run() method runs, then the notifyAll() will be called before the newly created thread calls wait(). The notification sent by the main thread is lost.

Henry