如果花费很长时间，python超时功能

Question

Let's say I have a function

def may_take_a_long_time():
    while 1:
        # do something
        if met_condition():
            break

timeout_dur = 600
may_take_a_long_time()
do_something_else()

What can I do so that if may_take_a_long_time() takes longer than timeout_dur , to abort whatever its doing, and proceed w/ do_something_else()

perhaps, setting SIGALARM before calling may_take_a_long_time() ? how do i do that? or run may_take_a_long_time() in a thread which happens async so i can have time.sleep(timeout_dur) in caller's thread?

Answer 1

Threading in python has quirks, and if you want an answer specific to that, I suggest you edit your question and put that up front.

For simple single threaded, you could use the existing timeout decorator as another user suggested or you can make your own, or you can do something like this (you'll have to tailor the code if you expect may_take_a_long_time() to return something, though:

import time

def may_take_a_long_time(tout):
    timeout = time.time() + tout
    passed = False

    while not passed:
        passed = met_condition()
        if time.time() > timeout:
            break

def met_condition():
   if some_condition:
      return True
   else:
      return False           

def do_something_else():
   # TODO your code here

if '__name__' == '__main__':
    may_take_a_long_time(30)
    do_something_else()