New submission from Sangeeth Saravanaraj: Python version 2.7.7 Mac OS Darwin Kernel Version 13.2.0
I have the following code which when executed waits to be interrupted by SIGINT, SIGTERM or SIGQUIT. When an object is initialized, it creates a threading.Condition() and acquires() it! The program then registers the signal handlers where notify() and release() is called when the above mentioned signals are received. After registering the signal handlers, it calls wait() on the condition variable and block. When I tried to stop the program with Ctrl-C, its did not respond. IOW, the _signal_handler() method did not get called. # start from signal import signal, SIGINT, SIGTERM, SIGQUIT from threading import Condition class A: def __init__(self): self._termination_signal = Condition() self._termination_signal.acquire(blocking=0) def _signal_handler(self, signum, frame): print "Received terminate request - signal = {0}".format(signum) del frame self._termination_signal.notify() self._termination_signal.release() return def register_and_wait(self): signal(SIGINT, self._signal_handler) signal(SIGTERM, self._signal_handler) signal(SIGQUIT, self._signal_handler) print "Waiting to be interrupted!" self._termination_signal.wait() # control blocks here! print "Notified!!" def main(): a = A() a.register_and_wait() if __name__ == "__main__": main() # end When the same code was tried in Python 3.4, it threw a "RuntimeError: cannot notify on un-acquired lock". More information is available in this conversation in python-list mailer - https://mail.python.org/pipermail/python-list/2014-July/674350.html ---------- components: Library (Lib) messages: 222250 nosy: pitrou, sangeeth priority: normal severity: normal status: open title: Possible deadlock in threading.Condition.wait() in Python 2.7.7 type: behavior versions: Python 2.7 _______________________________________ Python tracker <rep...@bugs.python.org> <http://bugs.python.org/issue21913> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com