Timeout Function with a Predefined Duration
In shell scripting, it's useful to have a mechanism to terminate long-running tasks to avoid excessive waiting. The challenge faced here is to wrap a function within a timeout script, returning False if it exceeds a specified time limit.
One approach is to set an asynchronous timer that triggers a False response after a pre-defined interval. Fortunately, such a solution is achievable using signal handlers in the signal library (available on UNIX-based systems).
The process involves creating a custom decorator (@timeout) that leverages signal.alarm() to set an alarm for the desired time interval. Within the decorated function, when the alarm expires, a TimeoutError exception is raised, effectively interrupting the operation.
To incorporate this solution into your code, save the following code as timeout.py and import it:
import errno import os import signal import functools 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
Now, within your application code, annotate any potentially long-running function with the @timeout decorator. For example:
from timeout import timeout # Short timeout @timeout(5) def slow_function(): # ... # Default timeout @timeout def another_slow_function(): # ... # Customize timeout and error message @timeout(30, error_message="Task timed out") def yet_another_slow_function(): # ...
This approach ensures that if a function takes longer than the specified timeout interval, it will raise a TimeoutError and return False, allowing you to handle unexpected delays gracefully.
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