multiprocessing PicklingError in Python 3.6.5 that doesn't occur in 3.6.1

Question:

The following code works in Python 3.6.1 … but returns this error in Python 3.6.5:

Traceback (most recent call last):
  File "G:/GOOD/Coding/Deepthroat/Deepthroat2/Bin/Testing/gh.py", line 36, in <module>
    loops.start()
  File "C:Program FilesPython36libmultiprocessingprocess.py", line 105, in start
    self._popen = self._Popen(self)
  File "C:Program FilesPython36libmultiprocessingcontext.py", line 223, in _Popen
    return _default_context.get_context().Process._Popen(process_obj)
  File "C:Program FilesPython36libmultiprocessingcontext.py", line 322, in _Popen
    return Popen(process_obj)
  File "C:Program FilesPython36libmultiprocessingpopen_spawn_win32.py", line 65, in __init__
    reduction.dump(process_obj, to_child)
  File "C:Program FilesPython36libmultiprocessingreduction.py", line 60, in dump
    ForkingPickler(file, protocol).dump(obj)
_pickle.PicklingError: Can't pickle <function loops at 0x0000019D3E30B598>: it's not the same object as __main__.loops

Here is the code:

from multiprocessing import Process
import time

def timeout(mm):
    for i in range(6):
        time.sleep(0.1)
    print('stop all bla loops now')

    mm.terminate()
    return


def loops():
    for i in range(5):
        time.sleep(0.1)
        print('loop1')

    billy = 'loop1 done'

    for i in range(5):
        time.sleep(0.1)
        print('loop2')

    billy = 'loop2 done'

    for i in range(5):
        time.sleep(0.1)
        print('loop3')

    billy = 'loop3 done'


billy = 'cool'

loops = Process(target=loops)
loops.start()

timeout = Process(target=timeout(loops))
timeout.start()
Asked By: Rhys

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Answers:

This is actually just an error by you because you are overwriting loops, which means it can’t be pickled (pickle just stores a reference to the function). If you say this error didn’t occur in python3.6.1, than probably there are test missing in python3.6.1

Answered By: MegaIng

If the target and variable both having same name like loops = Process(target=loops) then this error returns.

I could solve my problem using the explanations by Medalng and comments of Rhys in above answer. Re-iterated it again to help others as I could catch Rhys’s comment in third visit.

Answered By: Anand Prakash