Python 3.6.5 中的多处理 PicklingError 在 3.6.1 中没有发生

Question

以下代码在python 3.6.1 ...但在python 3.6.5中返回此错误：

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

这是代码：

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()

Answer 1

这实际上只是您的错误，因为您正在覆盖loops，这意味着它不能被腌制(Pickle只是存储对该功能的引用(。如果您说此错误在python3.6.1中没有发生，那么python3.6.1

Answer 2

如果目标和变量都具有与 loops = Process(target=loops)相同的名称，则此错误返回。

我可以在上述答案中使用Medalng的解释和Rhys的评论来解决我的问题。再次重申它以帮助他人，因为我可以在第三次访问中观看Rhys的评论。