Following this example in the ØMQ docs, I'm trying to create a simple receiver. The example uses infinite loop. Everything works just fine. However, on MS Windows, when I hit CTRL+C to raise KeyboardInterrupt, the loop does not break. It seems that recv()
method somehow ignores the exception. However, I'd love to exit the process by hiting CTRL+C instead of killing it. Is that possible?
In response to the @Cyclone's request, I suggest the following as a possible solution:
import signal
signal.signal(signal.SIGINT, signal.SIG_DFL);
# any pyzmq-related code, such as `reply = socket.recv()`
A zmq.Poller
object seems to help:
def poll_socket(socket, timetick = 100):
poller = zmq.Poller()
poller.register(socket, zmq.POLLIN)
# wait up to 100msec
try:
while True:
obj = dict(poller.poll(timetick))
if socket in obj and obj[socket] == zmq.POLLIN:
yield socket.recv()
except KeyboardInterrupt:
pass
# Escape while loop if there's a keyboard interrupt.
Then you can do things like:
for message in poll_socket(socket):
handle_message(message)
and the for-loop will automatically terminate on Ctrl-C. It looks like the translation from Ctrl-C to a Python KeyboardInterrupt only happens when the interpreter is active and Python has not yielded control to low-level C code; the pyzmq recv()
call apparently blocks while in low-level C code, so Python never gets a chance to issue the KeyboardInterrupt. But if you use zmq.Poller
then it will stop at a timeout and give the interpreter a chance to issue the KeyboardInterrupt after the timeout is complete.
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