Tim Peters added the comment: Richard, that's a strange argument ;-) Since, e.g., a BoundedSemaphore(1) is semantically equivalent to a mutex, it's like saying some_mutex.release() usually raises an exception if the mutex isn't held at the time - but maybe it won't. If the docs _say_ it's not reliable, fair enough. But they don't say that. When a thread gimmick is _intended_ to be probabilistic, the docs say so (e.g., Queue.qsize()).
Probably would have been better if a bounded=False optional argument to Semaphore.__init__() had been added, rather than creating a new BoundedSemaphore class. But, as is, it could easily be made reliable: 1. Change Semaphore's condition variable to use an RLock (instead of the current Lock). 2. Stuff the body of BoundedSemaphore.release() in a `with self._cond:` block. Curiously, there doesn't appear to be a test that BoundedSemaphore.release ever raises ValueError. So I'll readily agree that nobody ever took this seriously ;-) ---------- _______________________________________ Python tracker <rep...@bugs.python.org> <http://bugs.python.org/issue19158> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
