thread

[sweety sinha]

class ThreadA extends Thread{

	public synchronized void run(){
		for(int i=0;i<2;i++){
		try{
			sleep(5*1000);
		}catch(InterruptedException e){
		}
		System.out.println("thread name :" +Thread.currentThread().getName());
		}
	}
	ThreadA(String s){
			}
	ThreadA(){
	}
	
	public static void main(String [] args) {
			
			Thread t2 = new Thread();//1
			t2.start(); //1
			ThreadA tha = new ThreadA();//2
			tha.start();//2
			
			System.out.println("thread main name :" +tha.currentThread().getName());
			
}
}

when this code is run it gives answer as
thread main name :main
thread name :Thread-0
thread name :Thread-0


When i put 2 i place of 1 then the answer becomes

thread main name :main
thread name :Thread-1
thread name :Thread-1

So my question is why it is not showing both thread name i.e thread name :Thread-0 & thread name :Thread-1
i have also made run method synchronized.

thanks in advance

[Victor Fernandez]

When you run the code as it is you get this :

thread main name :main
thread name :Thread-1
thread name :Thread-1

The first line is the System.out.println statement you have at line 24 in your source code, it shows the current Thread which is main. Note that to be correct and avoid warnings must use Thread.currentThread() , because this method is static for class Thread.

The other two lines are generated by the System.out.println statement inside the for loop in the run method.

In this code you are launching 2 threads:
- 1 thread instance of Thread class, which doesn't print anything to the screen.
- 1 thread instance of ThreadA class

If you comment out the line 19 and 20 you get this:
thread main name :main
thread name :Thread-0
thread name :Thread-0

Because you are launching just one thread, a instance of ThreadA class.

[Samrat Kakustam]

Hi,
Check below :-

    Thread t2 = new Thread();//1   ---Change this to ThreadA t2 = new ThreadA();
            t2.start(); //1   
            ThreadA tha = new ThreadA();//2   
            tha.start();//2   
               
            System.out.println("thread main name :" +tha.currentThread().getName());   
  

There at line //1 you are just creating a thread that has no code in its run() method. As you have started that one first, it is given number Thread-0. Next you made a thread object of a class that extends Thread with code to be executed inside run() method. If you change the line as I mentioned above then you'll see what you have asked for.
Main point is the Thread class implemets Runnable interface but provides no code inside run() method.

Regards,
Samrat.

[sweety sinha]

thanks both of you for the answer.
now i got it