Subject: Re: [boost] [Interprocess::Semaphore] Deadlock on more producers - one consumer problem
Date: 2009-06-18 10:46:31
James Mansion wrote:
> Ion Gaztañaga wrote:
>> A semaphore post only increases a waiting thread if count was 0.
>> That's the behaviour of a semaphore.
> Not of Dijkstra's design it isn't. Conceptually a waiting consumer is
> immediately given the token and
> woken up, and further tokens will wake more consumers whether they are
> already waiting
> or not.
> Looks like a bug to me.
My point is that (http://en.wikipedia.org/wiki/Semaphore_(programming)):
P(Semaphore s) // Acquire Resource
wait until s > 0, then s := s-1;
/* Testing and decrementing s must be atomic to avoid race conditions */
V(Semaphore s) // Release Resource
s := s+1; /* must be atomic */
P can only wake a single thread, because the semaphore count represents
the free resource count. The original post said:
"the post() unlocks only one of the 4 producers"
and that's logical. If there are 4 waiting threads, you need 4 posts to
Boost list run by bdawes at acm.org, gregod at cs.rpi.edu, cpdaniel at pacbell.net, john at johnmaddock.co.uk