Doubt in spinner_async.py example #37

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1. antonmosin commented on Nov 24, 2023

   

   antonmosin

   on Nov 24, 2023

   Last edited by antonmosin

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   Hi @CarMoreno I'm just working through the chapter, let me test my thinking.

   Based on pp.711-712, the program is not stuck in the infinite loop because:

   1. The execution of the main() is blocked until asyncio.run() returns.
   2. asyncio.run() will stop as soon as the the supervisor() returns the result.
   3. The supervisor() is blocked until slow() is running. Once slow() finishes, the evaluation proceeds to the return statement.

   Thus, even though the spinner Task is not properly cancelled, the supervisor() returns the result which finishes the event loop that is controlled by the asyncio.run()

   We can test this theory based on a slightly modified example from the Python documentation

   import asyncio
   
   async def cancel_me():
       print('cancel_me(): before sleep')
   
       try:
           # Wait for 1 hour
           await asyncio.sleep(3600)
       except asyncio.CancelledError:
           print('cancel_me(): cancel sleep')
           raise
       finally:
           print('cancel_me(): after sleep')
   
   async def main():
       # Create a "cancel_me" Task
       task = asyncio.create_task(cancel_me())
   
       # Wait for 1 second
       await asyncio.sleep(1)
   
       #task.cancel()
   
   asyncio.run(main())
   
   # Expected output:
   #
   #     cancel_me(): before sleep
   #     cancel_me(): cancel sleep
   #     cancel_me(): after sleep
   #     main(): cancel_me is cancelled now
   

   Here I commented out the task.cancel() line, but the Task is still finalised by the asyncio.run() itself. From its docstring

   This function always creates a new event loop and closes it at the end.
      It should be used as a main entry point for asyncio programs, and should
      ideally only be called once.