CertPal #33 on Threads

[Vince Kennedy]

It asks for the output after the code is executed:

 public class test extends Thread {
	public static void main(String arhs[]) {
		Run1 r = new Run2();
		Run1 r2 = (Run2) r;
		Thread thread = new Thread(r);
		Thread thread2 = new Thread(r2);
		thread.start();
		thread2.start();
	}
}
class Run1 implements Runnable {
	int x;
	public void run() {
		try {
			System.out.println(++x);
		} finally {
			System.out.println(++x);
		}		
	}
	public int getX() {
		return x;
	}
}
class Run2 extends Run1{}

When I read through it at first it seemed there were two separate threads being started, and since Run was synchronized and x is a member variable (initialized to 0), the first call on start would return 1, and 2. Then when the second thread started, it would also print 1 and 2.
It was wrong, and the answer seems to be 1,2,3,4 - as if there is only one instance of x. But it's not a static variable, is it? So for x to be the same variable accessed by thread1 and thread2, would it have to be the same instance of a thread? And you can't start the same thread twice, correct?

So why is x holding it's value in this code?

Thanks.

[Neha Daga]

because the runnable object instance passed to both threads is same. If you had created a new runnable object and passed it to second thread then it would have been 1 2 1 2.

[Vince Kennedy]

Neha Daga wrote:because the runnable object instance passed to both threads is same. If you had created a new runnable object and passed it to second thread then it would have been 1 2 1 2.

So you can have two threads with the same instance of Runnable? i.e. the Run2 r2 = (Run2)r1; isn't two instances of the same thread, but two threads with the same instance of runnable? So calling start on each is legal?

[Ankit Garg]

Vince Kennedy wrote:So you can have two threads with the same instance of Runnable?

Yes

i.e. the Run2 r2 = (Run2)r; isn't two instances of the same thread, but two threads with the same instance of runnable? So calling start on each is legal?

The line you mentioned just creates a sort of alias to r. Neither does it create two instances of a thread, nor it creates two threads with the same runnable. It is just a plain assignment. The next two lines create two threads with the same runnable. You can simplify the code like this

Run1 r = new Run2();
Thread thread = new Thread(r);
Thread thread2 = new Thread(r);
thread.start();
thread2.start();

[Vince Kennedy]

Ankit Garg wrote:

i.e. the Run2 r2 = (Run2)r; isn't two instances of the same thread, but two threads with the same instance of runnable? So calling start on each is legal?

The line you mentioned just creates a sort of alias to r. Neither does it create two instances of a thread, nor it creates two threads with the same runnable. It is just a plain assignment.

Ok, but if this is just a simple assignment/alias statement, does that mean there are two references to the same thread? So how come you can start them both?

There are either two thread instances here or two references to one thread instance. If it's two thread instances, using one runnable then I suppose I understand why the output is 1 2 3 4. If it's two references to one thread, I don't understand how start can be called on the same thread through two references.

[Ankit Garg]

Ok, but if this is just a simple assignment/alias statement, does that mean there are two references to the same thread?

A Runnable is not a Thread. You need to understand the difference. This statement of yours is correct

If it's two thread instances, using one runnable then I suppose I understand why the output is 1 2 3 4.