如何腌制 ssl.SSLContext 对象

How to pickle a ssl.SSLContext object(如何腌制 ssl.SSLContext 对象)

问题描述

Windows 上的 Python 3.5，试试这些:

Python 3.5 on windows, try these:

import ssl, pickle, multiprocessing
context = ssl.create_default_context()
foo = pickle.dumps(context)
pickle.loads(foo)

抛出异常:

TypeError: __new__() missing 1 required positional argument: 'protocol'

multiprocessing.Process 的子类抛出同样的异常:

subclass of multiprocessing.Process throws the same exception:

class Foo(multiprocessing.Process):
    def __init__(self):
        super().__init__()
        self.context = ssl.create_default_context()

    def run(self):
        pass

if __name__ == '__main__':
    foo = Foo()
    foo.start()

推荐答案

这样的东西应该可以工作:

Something like this should work:

>>> import pickle, copyreg, ssl
>>>
>>> def save_sslcontext(obj):
...   return obj.__class__, (obj.protocol,)
... 
>>> copyreg.pickle(ssl.SSLContext, save_sslcontext)
>>> 
>>> context = ssl.create_default_context()
>>> foo = pickle.dumps(context)
>>> _foo = pickle.loads(foo)
>>> _foo
<ssl.SSLContext object at 0x1011812a8>
>>> _foo.protocol
2
>>> 

基本上，一个 SSLContext 需要一个 protocol，并且无论出于何种原因，protocol 都不会被保存(例如，它不在 >__reduce__ 方法)当实例被腌制时.如果您需要更多状态(即 __init__ 方法中的其他 args 和 kwds)，那么您需要从上面的 save_sslcontext 函数.(注意，如果你在 python 2.x 中，那么适当的模块是 copy_reg).

Basically, a SSLContext needs a protocol, and for whatever reason, the protocol is not saved (e.g. it's not in a __reduce__ method) when the instance is pickled. If you need more state (i.e. other args and kwds from the __init__ method), then you'll need to extend the return value from the save_sslcontext function above. (Note, if you are in python 2.x, then the appropriate module is copy_reg).

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