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

timeout-decorator don't work on windows system as , windows didn't support signal well.

If you use timeout-decorator in windows system you will get the following

AttributeError: module 'signal' has no attribute 'SIGALRM'

Some suggested to use use_signals=False but didn't worked for me.

Author @bitranox created the following package:

pip install https://github.com/bitranox/wrapt-timeout-decorator/archive/master.zip

Code Sample:

import time
from wrapt_timeout_decorator import *

@timeout(5)
def mytest(message):
    print(message)
    for i in range(1,10):
        time.sleep(1)
        print('{} seconds have passed'.format(i))

def main():
    mytest('starting')


if __name__ == '__main__':
    main()

Gives the following exception:

TimeoutError: Function mytest timed out after 5 seconds