Threads - Yielding

[Leonardo Silva]

RHE page 230 Ex.6

public class TestThread2 extends Thread {
public void run(){
System.out.println("Starting");
yield();
System.out.println("Done");
}
public static void main(String args []){
TestThread2 tt = new TestThread2();
tt.start();
}
}
I ran this code and confirmed boths lines from the run(); are printed.
I realized Calling yield(); only moves the running thread to the Ready state If there�s
a higher priority thread in the waiting pool. Can anyone confirm that??

[Jaikumar Nair]

According to me you are wrong.
My explanation:
Calling yield() moves the thread to the waiting state. After it moved to the ready state, if there are are no high priority threads waiting to be run, the the current thread will resume running.
Guys, correct my if I am wrong.
JAI

[ratul banji]

Yes..JOY..u r right.
Here , there is no 2nd thread ..that is why ..it is printing both.
Hope that helps.

[Naveen Sharma]

No Joi I Dont Agree With U. Yield methods just free all the resources and send the thread to the ready-to-run state and then its execution depends upon the operating system's thread scheduler.
Thanks

[Jane Griscti]

Hi all,
Naveen is correct. <code>yeild()</code> lets the scheduler know the thread is willing to let another thread run in it's place. In the end the scheduler will decide which thread gets control. It could be a higher priority thread, a thread with the same priority, one with lower priority or the original thread. The behaviour depends on how the scheduler is implemented on the system. There are no guarantees.
Leonardo, in your example, only one thread is active so <code>yield()</code> behaviour can't truly be seen. Try writing some code that creates a number of threads with different priorities and have them yeild. Of course, the outcome will depend on your platform and won't necessarily occur on all platforms.
Hope that helps.
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Jane Griscti
Sun Certified Programmer for the Java� 2 Platform