Re[4]: Re: New stackless proposal

On Sun, 11 Jul 2004 22:17:50 -0700 Mark Hahn <[email protected]> wrote:

> Lenard Lindstrom wrote:
> 
> > The question is whether or not prothon threads will be preemptive at
> > the byte code level? I do not know if this is the case for Stackless
> > tasklets so assumed they were. If the generator lacks a yield -
> > entirely possible - or fails to call it the thread calling the
> > generator may call its yield before the generator has completed. So
> > the calling thread will wait indefinitely.
> 
> Even if the calling thread has already hit its yield it can still get an
> exception and leave the yield.  There is no difference before and after
> getting to the yield.  You are saying there is a race but you are wrong.
> 
> > If the first StopIteration
> > is handled using yield then the calling thread is guaranteed to
> > return.
> 
> I don't understand what situation could cause the calling thread to not
> return.  Ultimately the generator thread must hit a yield or terminate.  So
> it will either yield a value or cause an exception.  There is no third
> possibility.
> 
So a terminating thread will raise an exception on all threads waiting on it?

def genfun():
    sleep(1.0)  # A very intensive computation
    # Return without calling yield
generator = Thread(genfun)

iter = generator().iter_()  # Generator goes to sleep
iter.next()                 # Now yielding to sleeping generator thread. Will I hang?

Lenard Lindstrom
<[email protected]>