How to return from function if got stuck for 90 seconds? [duplicate]

Question

Possible Duplicate:
Timeout on a Python function call

I want to implement that when the function took more than 90 seconds to complete it should return immediately when timeout. Is there any way to achieve that?

def abc(string):
    import re
    if re.match('some_pattern', string):
        return True
    else:
        return False

abc('some string to match')

Edited

Please download this test file. I have created a thread class and raise an exception within thread if timeout error occur. But thread is still alive because it prints i am still alive :) even after exception. Why an exception does not force the thread to stop??

Answer 1

I've edited my post to use jcollado's idea which is simpler.

The multiprocessing.Process.join method has a timeout argument which you can use like this:

import multiprocessing as mp
import time
import logging  
import re

logger = logging.getLogger(__name__)

def abc(string, result, wait = 0):
    time.sleep(wait)
    result.put(bool(re.match('some_pattern', string)))

if __name__ == '__main__':
    logging.basicConfig(level = logging.DEBUG,
                        format = '%(asctime)s:  %(message)s',
                        datefmt = '%H:%M:%S', )
    result = mp.Queue()
    proc = mp.Process(target = abc, args = ('some_pattern to match', result))
    proc.start()
    proc.join(timeout = 5)
    if proc.is_alive():
        proc.terminate()
    else:
        logger.info(result.get())

    proc = mp.Process(target = abc, args = ('some string to match', result, 20))
    proc.start()
    proc.join(timeout = 5)
    if proc.is_alive():
        logger.info('Timed out')
        proc.terminate()
    else:
        logger.info(result.get())

yields

12:07:59:  True
12:08:04:  Timed out

Note that you get the "Timed out" message in 5 seconds, even though abc('some string',20) would have taken around 20 seconds to complete.