How to signal alarm in python 2.4 after 0.5 seconds

Question

I want to timeout a particular piece of python code after in runs for 0.5 seconds. So I intend to raise an exception/signal after 0.5 seconds, and handle it gracefully and continue with rest of code.

In python i know that signal.alarm() can set alarm for integer seconds. Is there any alternative where we can generate an alarm after 0.5 seconds. signal.setitimer() as suggested in other posts is not available in python2.4 and I need to use python2.4 for this purpose?

Answer 1

Raise the alarm from a patiently waiting "daemon" thread. In the code below, snoozealarm does what you want via a SnoozeAlarm thread:

#! /usr/bin/env python

import os
import signal
import threading
import time

class SnoozeAlarm(threading.Thread):
  def __init__(self, zzz):
    threading.Thread.__init__(self)
    self.setDaemon(True)
    self.zzz = zzz

  def run(self):
    time.sleep(self.zzz)
    os.kill(os.getpid(), signal.SIGALRM)

def snoozealarm(i):
  SnoozeAlarm(i).start()

def main():
  snoozealarm(0.5)
  while True:
    time.sleep(0.05)
    print time.time()


if __name__ == '__main__':
  main()