Thread

[cyril vidal]

Dear all,
Question n�19 from Dan's (third part) devoted to threads:

 
class A extends Thread {
  private [B]static[/B] B b = new B();
  private String s1;
  public void run() {
    System.out.print(b.m1(s1));
  }
  A(String threadName, String s1) {
    super(threadName);
    this.s1 = s1;
  }
  public static void main (String[] args) {
    A a = new A("T1","A");
    A b = new A("T2","B");
    a.start();
    b.start();
  }
}
class B {
  private String s1;
  public  String m1(String s) {
    s1 = s;
    try {Thread.sleep(1);} catch (InterruptedException ie) {}
    return "[" +
           Thread.currentThread().getName() +
           "," + s1 + "]";
  }
}
 

Answer for possible outputs:

Prints: [T1,A][T2,B].
c. Prints: [T1,B][T2,B].
d. Prints: [T2,B][T1,A].
e. Prints: [T2,A][T1,A].
Since method m1 is not synchronized

As far as I understand, here the modifier static for object b in class A is very important: it means that the same object B is shared between the two threads a and b.
But if we delete this keyword in the code, now each thread has its own object b, and there is no more reason that we obtain a different result than

[T1,A][T2,B]

I've tested it, and on my computer, that's indeed the case.
But may we generalize?
Thanks in advance for your answer,
Cyril.

[Miki Muzsi]

If you remove the static keyword, indeed there are two diffeerent instances of B created. This garantees that you can not get output like:
[T1,B]
or [T2,A]
However you will not have garantee that the output is always: [T1,A][T2,B]. It can be [T2,B] [T1,A] as well, since no control (garantee) how thread scheduler will do things.
Miki

[cyril vidal]

That's right!
Thanks Miki for your answer.