Q from Whizlabs

[Claudio Roitman]

Hi, Mock Exam practice number 9 from Whizlabs 4.0, Question number 27 says:
"Which of the following statements are true?
a) A Thread's start() method, automatically call the run() method
b)...
c)...
d)...
"
It says option a is true, But I think it is not, because start() method does not call run() method, It just put the receiver thread in a runnable state, and then, The Thread scheduler will decide when to invoke it's run() method...(maybe never!)
Am I rigth?
Thanks,
Claudio

[fei peng]

Hi Claudio,
I think your understand is right,only the scheduler can decide which thread to run.
but this answer doesnt care whether the run() MUST be run,so this issue has no problem

[Marlene Miller]

Hi Claudio, I agree with you.
The Java Programming Language, section 10.1 Creating Threads says "The start method spawns a new thread of control based on the data in the Thread object, then returns. Now the virtual machine invokes the new thread's run method, making the thread active."
I wonder if what Whizlabs was trying to say is that the client calls start() but does not call run().
[ November 28, 2003: Message edited by: Marlene Miller ]

[Mona Gadkari]

Originally posted by Marlene Miller:
Hi Claudio, I agree with you.
I wonder if what Whizlabs was trying to say is that the client calls start() but does not call run().
[ November 28, 2003: Message edited by: Marlene Miller ]

Hello Marlene
Now 'am in a fix what r we supposed to mark in the exam. I always thought that that the start() calls the run(). Coz i had read somewhere that we do not directly call the run().

[Marlene Miller]

Hi Mona,

Now 'am in a fix what r we supposed to mark in the exam.

On the exam, I would say start() does not call run().

I always thought that that the start() calls the run(). Coz i had read somewhere that we do not directly call the run().

We do not know for sure what happens inside start(), because we do not have the source code. start() is a native method.
public synchronized native void start();
----
I assume start() does not call run() for the following reason. Suppose the main() method calls start(). That is, the main thread executes start(). If start() called run(), the main thread would execute run(). But we know for sure the main thread does not execute run(). Instead the new thread executes run().
Two threads execute two different paths through the code. One thread executes start(), the other thread executes run(). Sometimes the virtual machine executes one path, sometimes the other path.
----
I believe the book The Java Programming Language is trustworthy. One of the coauthors is James Gosling, the creator of the Java language.
"The start method spawns a new thread of control based on the data in the Thread object, then returns. Now the virtual machine invokes the new thread's run method, making the thread active."
----
We can almost prove start() does not call run().

class Test extends Thread {
   public void run() {
      System.out.print("run()");
   }
   public static void main(String[] args) {
      Test t1 = new Test();
      t1.start();
      System.out.print("main()");
   }
}
Output:
main()run()
//
class Test extends Thread {
   public void run() {
      System.out.print("run()");
   }
   public static void main(String[] args) {
      Test t1 = new Test();
      t1.setPriority(10);
      t1.start();
      System.out.print("main()");
   }
}
Output:
run()main()

The first time, it looks like start() returns and then the virtual machine executes run(). The second time, I increased the priority of the new thread. It looks like the virtual machine executes run() and then start() returns.

[Marlene Miller]

Thread-1                    heap
  | main()
  |   t1=new Thread()
  |                        --------
  |   t1 ---------------> | Thread |
  |                +----> | object |<---+
  |                |       --------     |
  |   t1.start()   |                    | 
  |     this------>+                    |
  |                                     |
  |                                     |   Thread-2
  |                                     |     run()
  |                                     +<------this      

[ November 30, 2003: Message edited by: Marlene Miller ]