为什么多进程在 ubuntu 和 macOS 中的工作方式不同？

Question

我有这个简单的代码来使事情变得更简单。

from multiprocessing import Process
def abc():
    print("HI")
    print(a)

a = 4
p = Process(target = abc)
p.start()

它在 ubuntu (Python 3.8.5) 中运行得很好，并提供输出：

HI
4

但是，它在spyder（Python 3.9.5）“AttributeError：无法在<模块' main '（内置）>上获取属性'abc'”和macOS（Python 3.8.10，我也尝试了其他版本）中失败并失败）CLI“运行时错误”。

间谍错误：

Traceback (most recent call last):
  File "<string>", line 1, in <module>
  File "multiprocessing/spawn.pyc", line 116, in spawn_main
  File "multiprocessing/spawn.pyc", line 126, in _main
AttributeError: Can't get attribute 'abc' on <module '__main__' (built-in)>
Traceback (most recent call last):
  File "<string>", line 1, in <module>
  File "multiprocessing/spawn.pyc", line 116, in spawn_main
  File "multiprocessing/spawn.pyc", line 126, in _main
AttributeError: Can't get attribute 'abc' on <module '__main__' (built-in)>

MacOS-BigSur 错误：

Traceback (most recent call last):
  File "<string>", line 1, in <module>
  File "/Users/asavas/opt/anaconda3/lib/python3.8/multiprocessing/spawn.py", line 116, in spawn_main
    exitcode = _main(fd, parent_sentinel)
  File "/Users/asavas/opt/anaconda3/lib/python3.8/multiprocessing/spawn.py", line 125, in _main
    prepare(preparation_data)
  File "/Users/asavas/opt/anaconda3/lib/python3.8/multiprocessing/spawn.py", line 236, in prepare
    _fixup_main_from_path(data['init_main_from_path'])
  File "/Users/asavas/opt/anaconda3/lib/python3.8/multiprocessing/spawn.py", line 287, in _fixup_main_from_path
    main_content = runpy.run_path(main_path,
  File "/Users/asavas/opt/anaconda3/lib/python3.8/runpy.py", line 265, in run_path
    return _run_module_code(code, init_globals, run_name,
  File "/Users/asavas/opt/anaconda3/lib/python3.8/runpy.py", line 97, in _run_module_code
    _run_code(code, mod_globals, init_globals,
  File "/Users/asavas/opt/anaconda3/lib/python3.8/runpy.py", line 87, in _run_code
    exec(code, run_globals)
  File "/Users/asavas/delete.py", line 8, in <module>
    p.start()
  File "/Users/asavas/opt/anaconda3/lib/python3.8/multiprocessing/process.py", line 121, in start
    self._popen = self._Popen(self)
  File "/Users/asavas/opt/anaconda3/lib/python3.8/multiprocessing/context.py", line 224, in _Popen
    return _default_context.get_context().Process._Popen(process_obj)
  File "/Users/asavas/opt/anaconda3/lib/python3.8/multiprocessing/context.py", line 284, in _Popen
    return Popen(process_obj)
  File "/Users/asavas/opt/anaconda3/lib/python3.8/multiprocessing/popen_spawn_posix.py", line 32, in __init__
    super().__init__(process_obj)
  File "/Users/asavas/opt/anaconda3/lib/python3.8/multiprocessing/popen_fork.py", line 19, in __init__
    self._launch(process_obj)
  File "/Users/asavas/opt/anaconda3/lib/python3.8/multiprocessing/popen_spawn_posix.py", line 42, in _launch
    prep_data = spawn.get_preparation_data(process_obj._name)
  File "/Users/asavas/opt/anaconda3/lib/python3.8/multiprocessing/spawn.py", line 154, in get_preparation_data
    _check_not_importing_main()
  File "/Users/asavas/opt/anaconda3/lib/python3.8/multiprocessing/spawn.py", line 134, in _check_not_importing_main
    raise RuntimeError('''
RuntimeError: 
        An attempt has been made to start a new process before the
        current process has finished its bootstrapping phase.

        This probably means that you are not using fork to start your
        child processes and you have forgotten to use the proper idiom
        in the main module:

            if __name__ == '__main__':
                freeze_support()
                ...

        The "freeze_support()" line can be omitted if the program
        is not going to be frozen to produce an executable.

试图了解为什么以及如何解决这个问题？

谢谢

Answer 1

看来这个版本的 macOS 已改用使用spawn而不是fork创建新进程的方法。这意味着当创建一个新进程时，会创建一个新的空地址空间，启动一个新的Python解释器，并从顶部重新执行源文件。因此，任何处于全局范围且不在开始的块内的代码都if __name__ == '__main__':将被执行。这就是为什么创建进程的任何代码都必须位于这样的块中，否则您将陷入递归的进程创建循环，从而生成您看到的错误。您只需要：

from multiprocessing import Process

def abc():
    print("HI")
    print(a)

# this will get re-executed in the new subprocess:
a = 4

if __name__ == '__main__':
    p = Process(target=abc)
    p.start()
    p.join() # explicitly wait for process to complete