Forgive me, I am a newbie. I've surveyed some solution. But it is so hard for me to understand and to modify that. (Or maybe there is no solution in line with my imagination?). And I hope it can work on Ubuntu & Win7.
There is an example like this.
import random,time
def example():
while random.randint(0,10) != 1:
time.sleep(1)
print "down"
example()
And my imagination is...
If, the example() run over 10s, then rerun the example() again. (And maybe there is a place I can code anything else. Like I want to record the timeout event on TXT, and I can code the code at that place.) Else, do nothing.
Is it possible to do that?
You can run a watch-dog in a separate thread that interrupts the main thread (that runs example) when it exceeds the time limit. Here is a possible implementation, with timeout lowered to 3s for ease of debugging:
import time, threading, thread
def watchdog_timer(state):
time.sleep(3)
if not state['completed']:
thread.interrupt_main()
def run_example():
while True:
state = {'completed': False}
watchdog = threading.Thread(target=watchdog_timer, args=(state,))
watchdog.daemon = True
watchdog.start()
try:
example()
state['completed'] = True
except KeyboardInterrupt:
# this would be the place to log the timeout event
pass
else:
break
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