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Subclassing A Class With Private Members

- 1 answer

One of the really nice things about python is the simplicity with which you can name variables that have the same name as the accessor:

self.__value = 1

def value():
    return self.__value

Is there a simple way of providing access to the private members of a class that I wish to subclass? Often I wish to simply work with the raw data objects inside of a class without having to use accessors and mutators all the time.

I know this seems to go against the general idea of private and public, but usually the class I am trying to subclass is one of my own which I am quite happy to expose the members from to a subclass but not to an instance of that class. Is there a clean way of providing this distinction?

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Answer

Not conveniently, without further breaking encapsulation. The double-underscore attribute is name-mangled by prepending '_ClassName' for the class it is being accessed in. So, if you have a 'ContainerThing' class that has a '__value' attribute, the attribute is actually being stored as '_ContainerThing__value'. Changing the class name (or refactoring where the attribute is assigned to) would mean breaking all subclasses that try to access that attribute.

This is exactly why the double-underscore name-mangling (which is not really "private", just "inconvenient") is a bad idea to use. Just use a single leading underscore. Everyone will know not to touch your 'private' attribute and you will still be able to access it in subclasses and other situations where it's darned handy. The name-mangling of double-underscore attributes is useful only to avoid name-clashes for attributes that are truly specific to a particular class, which is extremely rare. It provides no extra 'security' since even the name-mangled attributes are trivially accessible.

For the record, '__value' and 'value' (and '_value') are not the same name. The underscores are part of the name.

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source: stackoverflow.com
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