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

Here is a slight improvement to the given thread-based solution.

The code below supports exceptions:

def runFunctionCatchExceptions(func, *args, **kwargs):
    try:
        result = func(*args, **kwargs)
    except Exception, message:
        return ["exception", message]

    return ["RESULT", result]


def runFunctionWithTimeout(func, args=(), kwargs={}, timeout_duration=10, default=None):
    import threading
    class InterruptableThread(threading.Thread):
        def __init__(self):
            threading.Thread.__init__(self)
            self.result = default
        def run(self):
            self.result = runFunctionCatchExceptions(func, *args, **kwargs)
    it = InterruptableThread()
    it.start()
    it.join(timeout_duration)
    if it.isAlive():
        return default

    if it.result[0] == "exception":
        raise it.result[1]

    return it.result[1]

Invoking it with a 5 second timeout:

result = timeout(remote_calculate, (myarg,), timeout_duration=5)