How does this watchdog function work?

Asked Mar 23 '17 at 14:40

Active Oct 18 '17 at 10:02

Viewed 733 times

This is from an example found on this site. I am not clear on when the Watchdog function is executed in the except clause. In my opinion this will never be executed unless there is an error. What am I missing?

from threading import Timer
import time
class Watchdog:
    def __init__(self, timeout, userHandler=None):  # timeout in seconds
        self.timeout = timeout
        self.handler = userHandler if userHandler is not None else self.defaultHandler
        self.timer = Timer(self.timeout, self.handler)

    def reset(self):
        self.timer.cancel()
        self.timer = Timer(self.timeout, self.handler)

    def stop(self):
        self.timer.cancel()

    def defaultHandler(self):
        raise self


watchdog = Watchdog(2)
watchdog.timer.start()
x=1
try:
  # do something that might take too long
  while x>0:
    print "test"
    time.sleep(0.2)
except watchdog:
  # handle watchdog error
  watchdog.stop()

edited Oct 18 '17 at 10:02

Maxim

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asked Mar 23 '17 at 14:40

MiniMe

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It's basically a trick for terminating code that is taking a long time to execute. – pzp Mar 23 '17 at 14:45
Yeah but it doesn't seem to work – MiniMe Mar 23 '17 at 14:50
it seems `self.timer.start()` is missing – Paweł Kordowski Mar 23 '17 at 14:55
where do I place that? Placing that just after instantiating Watchdog did not work. It did rise an error but it did not break the loop – MiniMe Mar 23 '17 at 15:12
Could you give the reference of where you found that code. It seems to be incomplete – Serge Ballesta Mar 23 '17 at 15:15
@Serge-Ballesta http://stackoverflow.com/questions/16148735/how-to-implement-a-watchdog-timer-in-python – MiniMe Mar 23 '17 at 16:49

How does this watchdog function work?

0 Answers0