How to Interrupt a Python Function Execution with a Timeout?

Interrupting a Function with Timeout in Python

When calling functions that may stall indefinitely, preventing the script from executing further, it becomes necessary to implement a timeout mechanism. Python's signal package provides a solution for this issue.

The signal package, primarily used in UNIX systems, allows you to set up a timeout for a specific function. If the function exceeds the specified timeout, a signal is raised to interrupt the execution.

Example:

Consider a function loop_forever() that may run indefinitely. We need to call this function but set a timeout of 5 seconds. If the function takes longer than 5 seconds, we want to cancel its execution.

import signal

# Register a handler for the timeout
def handler(signum, frame):
    print("Timeout! Cancelling function execution.")
    raise Exception("Timeout exceeded!")

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

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

try:
    loop_forever()
except Exception as e:
    print(str(e))
    
# Cancel the timer if the function finishes before timeout
signal.alarm(0)

In this example, after 5 seconds, the handler function is called, raising an exception. This exception is caught in the parent code, which then cancels the timer and terminates the execution of the loop_forever() function.

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