threads

[ushma agarwal]

please tell me why the output of the following code is T1T1T3.

  
class A implements Runnable {
  public void run() {System.out.print(Thread.currentThread().getName());}
}
class B implements Runnable {
  public void run() {
    new A().run();
    new Thread(new A(),"T2").run();
    new Thread(new A(),"T3").start();
}}
class C {
  public static void main (String[] args) {
    new Thread(new B(),"T1").start();
}}
  

why not T1T2T3.

new Thread(new A(),"T2").run();
here we are creating a new thread "T2" and on this thread we are invoking run() method directly.

[Kelvin Chenhao Lim]

Invoking the run() method doesn't start a new thread. All you're doing is making a normal method invocation, i.e. you're executing the statements in run() within the same thread that called it. You need to use the Thread object's start() method to actually create a new thread. start() will then invoke run() for you in the new thread.

(This is an extremely common point of confusion, by the way. I can't even begin to estimate the number of times I've spotted junior programmers making this mistake at my workplace.)

[nico dotti]

As Kelvin says directly calling run() will not take you into a new thread's execution context (T2) so it's still T1. Look at the code and see where the direct call to run() is made, think about what the current thread is at that point, and note that THAT will be the current thread when the run() method is executed (just as if you called any other method like foo() or bar(). In fact, when you created the T2 Thread in went into the 'NEW' state because you did get an instance of Thread, but the only way it will go into the 'RUNNABLE' state, and eventually on to RUNNING, is if you call start(). Not saying anything different than Kelvin really but maybe it'll help seeing it explained two different ways.
[ December 07, 2007: Message edited by: nico dotti ]

[ushma agarwal]

Thanks a lot Kelvin and nico for explaining it so nicely