python - 如何在Python中实现看门狗计时器？

我想用两个用例在Python中实现一个简单的看门狗计时器:

看门狗确保函数的执行时间不超过x秒

看门狗确保某些定期执行的函数确实至少每隔y秒执行一次

我怎么做？

最佳答案

只需发布我自己的解决方案:

from threading import Timer

class Watchdog(Exception):
    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)
        self.timer.start()

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

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

    def defaultHandler(self):
        raise self

如果要确保函数在不到x秒内完成，请使用:

watchdog = Watchdog(x)
try:
  # do something that might take too long
except Watchdog:
  # handle watchdog error
watchdog.stop()

如果您定期执行某件事并希望确保至少每隔y秒执行一次，则使用此方法:

import sys

def myHandler():
  print "Whoa! Watchdog expired. Holy heavens!"
  sys.exit()

watchdog = Watchdog(y, myHandler)

def doSomethingRegularly():
  # make sure you do not return in here or call watchdog.reset() before returning
  watchdog.reset()

关于python - 如何在Python中实现看门狗计时器？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/16148735/