On 3/14/06, Guido van Rossum <guido at python.org> wrote:
> Isn't this a job for threading.BoundedSpemaphore()?
Not sure I see how. What I think Raymond's after (and certainly what I
want) is to queue N tasks, set a counter to N, then wait until the
counter goes to zero.
I suppose
counter = Semaphore(-N)
# Queue N tasks
counter.acquire()
with each task (or the queue) saying
counter.release()
when it finishes. But the logic seems backwards, and highly prone to
off-by-one errors, and I'm not entirely convinced that a negative
semaphore value is officially supported...
(BoundedSemaphore seems a red herring here - the blocking semantics of
Semaphore and BoundedSemaphore are identical).
Paul.
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