Timeout on a function call

Question

I\'m calling a function in Python which I know may stall and force me to restart the script.

How do I call the function or what do I wrap it in so that if it takes

Answer 1

Building on and and enhancing the answer by @piro , you can build a contextmanager. This allows for very readable code which will disable the alaram signal after a successful run (sets signal.alarm(0))

@contextmanager
def timeout(duration):
    def timeout_handler(signum, frame):
        raise Exception(f'block timedout after {duration} seconds')
    signal.signal(signal.SIGALRM, timeout_handler)
    signal.alarm(duration)
    yield
    signal.alarm(0)

def sleeper(duration):
    time.sleep(duration)
    print('finished')

Example usage:

In [19]: with timeout(2):
    ...:     sleeper(1)
    ...:     
finished

In [20]: with timeout(2):
    ...:     sleeper(3)
    ...:         
---------------------------------------------------------------------------
Exception                                 Traceback (most recent call last)
 in ()
      1 with timeout(2):
----> 2     sleeper(3)
      3 

 in sleeper(t)
      1 def sleeper(t):
----> 2     time.sleep(t)
      3     print('finished')
      4 

 in timeout_handler(signum, frame)
      2 def timeout(duration):
      3     def timeout_handler(signum, frame):
----> 4         raise Exception(f'block timedout after {duration} seconds')
      5     signal.signal(signal.SIGALRM, timeout_handler)
      6     signal.alarm(duration)

Exception: block timedout after 2 seconds