regarding join() method in Thread

[Anish Nagraj]

public class Joining {

static Thread createThread(final int i, final Thread t1){

Thread t2 = new Thread(){
public void run(){
System.out.println(i+1);
try {
t1.join();
} catch (InterruptedException e) {
e.printStackTrace();
}
System.out.println(i+2);
}
};

System.out.println(i+3);
t2.start();
System.out.println(i+4);
return t2;
}

public static void main(String[] args) {

createThread(10, createThread(20, Thread.currentThread()));
}
}

can any body explain me the flow of this program?
What does join() actually do here?
Does the static specified for the method 'Joining' makes any change? Because I get different answers if and if not static is present
Plz help me...

[Vijay Gade]

Well Well...

public class Joining {
    static Thread createThread(final int i,final Thread t1){
        Thread t2 = new Thread(){
            public void run(){
                System.out.println(i+1);
                try {
                    t1.join();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println(i+2);
            }
        };
        System.out.println(i+3);
        t2.start();
        System.out.println(i+4);
        return t2;
    }
    public static void main(String[] args) {
        createThread(10, createThread(20, Thread.currentThread()));
    }
}

(I just came up with some kind of a solution based on the output :-). I am not a 100% sure, but I think I am close)
So think of it like this:
One thing is, the inner method gets executed first. And, I am shortening the name of the method createThread to just "CT", and I am giving the current thread a name, which M (as in main).

CT (10, CT (20, M))
                            |
CT (20,M) // This is what gets executed first
                            |
23 // As per the code
                            |
//The thread t2 is "scheduled" to start
                            |
24 // prints by the time the run method of t2 is invoked
                            |
21 // prints because by now, the run method has been invoked
                            |
//Now, t2 is instructed to join M. So it has to wait for M to finish
                            |
//But hey, t2 is now returned. So, since t2 is waiting on M to finish, and now M is returned, the number 22 can be printed any time now. It's upto the JVM implementation to decide when to print 22.
                            |
//Now, let's say the entire process that has been described before as a thread, and let's give it a name. M2 is going to be it for now!
                            |
//CT (10, M2) is now called, becuase CT (20, M) has finished it's job.
                            |
//Again, as per the code, 13 get's printed
                            |
//Schedule a new therad M2t2
                            |
//Again, by this time, 14 get's printed
                            |
//I saw 22 get printed now in my compiler. Try your's and see when 22 is printed. So by now, t2 is done. It waited on M (by "joining" M), and when M had been returned, it was upto the JVM to print 22, and now is when it chose
                            |
//and of course, if you have come this far, you would now understand that 11 and 12 are the numbers that are going to be printed next.

Basically, this was the ouput when I ran your program. And please make sure you use the Instant UBB code below when you paste your CODE.

23
24
21
13
14
22
11
12

Hope that helps.

Thanks,
-Vijay

[Vijay Gade]

The other answer to your question about being static or not is that, I am a little confused. Basically the createThread method was named static so that you can call it within your main method. If you remove the word 'static', your code would not even compile.

Thanks,
-Vijay

[Vidhya Hari]

This is one of those clearly unguaranteed behaviour of the JVM,if im right!