[Timeout a Long Running Function] #python3

timeout_context.py

import signal
import time

class timeout:
    def __init__(self, seconds=1, error_message='Timeout'):
        self.seconds = seconds
        self.error_message = error_message
    def handle_timeout(self, signum, frame):
        raise TimeoutError(self.error_message)
    def __enter__(self):
        signal.signal(signal.SIGALRM, self.handle_timeout)
        signal.alarm(self.seconds)
    def __exit__(self, type, value, traceback):
        signal.alarm(0)
        
if __name__ == "__main__":
	with timeout(seconds=3):
		time.sleep(4)

timeout_decorator.py

import errno
import os
import signal
import functools
import time

class TimeoutError(Exception):
    pass

def timeout(seconds=10, error_message=os.strerror(errno.ETIME)):
    def decorator(func):
        def _handle_timeout(signum, frame):
            raise TimeoutError(error_message)

        @functools.wraps(func)
        def wrapper(*args, **kwargs):
            signal.signal(signal.SIGALRM, _handle_timeout)
            signal.alarm(seconds)
            try:
                result = func(*args, **kwargs)
            finally:
                signal.alarm(0)
            return result

        return wrapper

    return decorator

# Timeout a long running function with the default expiry of 10 seconds.
@timeout
def long_running_function1(secs:int) -> None:
	time.sleep(secs)

# Timeout after 5 seconds
@timeout(5)
def long_running_function2(secs:int) -> None:
	time.sleep(secs)

# Timeout after 30 seconds, with the error "Connection timed out"
@timeout(30, os.strerror(errno.ETIMEDOUT))
def long_running_function3(secs:int) -> None:
	time.sleep(secs)
  
if __name__ == "__main__":
	long_running_function1(11)
	long_running_function2(6)
	long_running_function3(31)