为什么ubuntu和macOS中的多进程工作方式不同？

Question

我有这个简单的代码使事情变得更简单

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）中运行良好，并提供以下输出：

HI
4

但是，它在spyder（Python 3.9.5）“AttributeError:无法在<；模块'main'（内置）>；“和macOS（Python 3.8.10，我也尝试了其他版本）和CLI“RuntimeError”上获取属性“abc”

Spyder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.

试图了解我为什么以及如何解决此问题

谢谢

Answer 1

这个版本的macOS似乎已经切换到使用spawn而不是fork方法来创建新进程。这意味着当创建一个新进程时，将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，在全局范围内且不在以if __name__ == '__main__':开头的块内的任何代码都将被执行。这就是为什么任何创建进程的代码都必须在这样一个块中，否则您将进入一个递归的进程创建循环，该循环将生成您看到的错误。你只需要：

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