How to Handle Function Timeouts in Python?

Handling Function Timeouts in Python

When calling a function that may potentially stall, it becomes crucial to have a mechanism in place to gracefully handle such situations and prevent the script from freezing indefinitely. One effective solution to this problem is to set a timeout for the function call.

The Python signal module provides the necessary functionality to register a signal handler that will be invoked if the function exceeds the specified timeout. Here's how it can be implemented:

import signal

def handler(signum, frame):
    print("Timeout reached!")
    raise Exception("Function call timed out")

def function_to_timeout():
    print("Running function...")
    # Simulate a long-running task
    for i in range(10):
        time.sleep(1)

# Register the signal handler
signal.signal(signal.SIGALRM, handler)

# Set a timeout of 5 seconds
signal.alarm(5)

try:
    # Call the function
    function_to_timeout()
except Exception as exc:
    print(exc)

When the function_to_timeout() is called, it will print a message indicating that it is running. The signal.alarm(5) sets a 5-second timeout, after which the handler function will be triggered. Within the handler, an exception is raised to terminate the function call.

This approach allows you to set a specific timeout for the function and gracefully exit the script in case the function takes longer than the specified time.

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