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

How do I call the function or what do I wrap it in so that if it takes longer than 5 seconds the script cancels it?

I posted a gist that solves this question/problem with a decorator and a threading.Timer. Here it is with a breakdown.

Imports and setups for compatibility

It was tested with Python 2 and 3. It should also work under Unix/Linux and Windows.

First the imports. These attempt to keep the code consistent regardless of the Python version:

from __future__ import print_function
import sys
import threading
from time import sleep
try:
    import thread
except ImportError:
    import _thread as thread

Use version independent code:

try:
    range, _print = xrange, print
    def print(*args, **kwargs): 
        flush = kwargs.pop('flush', False)
        _print(*args, **kwargs)
        if flush:
            kwargs.get('file', sys.stdout).flush()            
except NameError:
    pass

Now we have imported our functionality from the standard library.

exit_after decorator

Next we need a function to terminate the main() from the child thread:

def quit_function(fn_name):
    # print to stderr, unbuffered in Python 2.
    print('{0} took too long'.format(fn_name), file=sys.stderr)
    sys.stderr.flush() # Python 3 stderr is likely buffered.
    thread.interrupt_main() # raises KeyboardInterrupt

And here is the decorator itself:

def exit_after(s):
    '''
    use as decorator to exit process if 
    function takes longer than s seconds
    '''
    def outer(fn):
        def inner(*args, **kwargs):
            timer = threading.Timer(s, quit_function, args=[fn.__name__])
            timer.start()
            try:
                result = fn(*args, **kwargs)
            finally:
                timer.cancel()
            return result
        return inner
    return outer

Usage

And here's the usage that directly answers your question about exiting after 5 seconds!:

@exit_after(5)
def countdown(n):
    print('countdown started', flush=True)
    for i in range(n, -1, -1):
        print(i, end=', ', flush=True)
        sleep(1)
    print('countdown finished')

Demo:

>>> countdown(3)
countdown started
3, 2, 1, 0, countdown finished
>>> countdown(10)
countdown started
10, 9, 8, 7, 6, countdown took too long
Traceback (most recent call last):
  File "", line 1, in 
  File "", line 11, in inner
  File "", line 6, in countdown
KeyboardInterrupt

The second function call will not finish, instead the process should exit with a traceback!

KeyboardInterrupt does not always stop a sleeping thread

Note that sleep will not always be interrupted by a keyboard interrupt, on Python 2 on Windows, e.g.:

@exit_after(1)
def sleep10():
    sleep(10)
    print('slept 10 seconds')

>>> sleep10()
sleep10 took too long         # Note that it hangs here about 9 more seconds
Traceback (most recent call last):
  File "", line 1, in 
  File "", line 11, in inner
  File "", line 3, in sleep10
KeyboardInterrupt

nor is it likely to interrupt code running in extensions unless it explicitly checks for PyErr_CheckSignals(), see Cython, Python and KeyboardInterrupt ignored

I would avoid sleeping a thread more than a second, in any case - that's an eon in processor time.

How do I call the function or what do I wrap it in so that if it takes longer than 5 seconds the script cancels it and does something else?

To catch it and do something else, you can catch the KeyboardInterrupt.

>>> try:
...     countdown(10)
... except KeyboardInterrupt:
...     print('do something else')
... 
countdown started
10, 9, 8, 7, 6, countdown took too long
do something else