Python - Redefining class attribute in a derived class

Question

Suppose I have a base class, where there's a __flag attribute accessible by a @classmethod.

class Base(object):
    __flag = None

    def __init__(self) :
        pass

    @classmethod
    def flag(self):
        return self.__flag

Suppose I also have a derived class where I change the attribute.

class Derived(Base):
    __flag = True

Then, when I try to access the attribute of the derived class, I get the attribute of the base class:

In [3]: print Derived.flag()
None

Why? I really can not understand.

Wolph · Accepted Answer · 2013-10-15 09:16:29Z

This is because the "hidden" variables in Python get stored differently. There's a tiny bit of magic there.

Here's an example of why it does not work:

class Base(object):
    __flag = 'base'
    _other_flag = 'base'

    def __init__(self) :
        pass

    @classmethod
    def flag(self):
        return self.__flag

    @classmethod
    def other_flag(self):
        return self._other_flag

class Derived(Base):
    __flag = 'derived'
    _other_flag = 'derived'

print 'base flag', Base.flag()
print 'derived flag', Derived.flag()
print 'base other flag', Base.other_flag()
print 'derived other flag', Derived.other_flag()

# Note the following 2 statements:
print 'base flag property', Derived._Base__flag
print 'derived flag property', Derived._Derived__flag

print 'base other flag property', Base._other_flag
print 'derived other flag property', Derived._other_flag

As you can see at the bottom, it's stored in a different variable and silently translated to that within the Base.flag method.

Aleksander Lidtke · Accepted Answer · 2013-10-15 09:12:45Z

2

Remove the __ from __flag and it will work nicely. Like so:

class Base(object):
    flagAtt = None

    def __init__(self) :
       pass

    @classmethod
    def flag(self):
        return self.flagAtt

class Derived(Base):
    flagAtt = True

Which gives:

>>> print Derived.flag()
True
>>> print Base.flag()
None

answered Oct 15, 2013 at 9:12

Aleksander Lidtke

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1 Comment

Andrey Sobolev Over a year ago

Oh, it's as simple as that! Thank you very much!

uranusjr · Accepted Answer · 2013-10-15 09:17:43Z

1

Python performs name-mangling on variables that start with two underscores (except those that also end with two underscores), so __flag in Base is different from __flag. You can observe this by using dir:

>>> class Base(object):
...     __flag = True
... 
>>> dir(Base)
['_Base__flag', '__class__', '__delattr__', ......   # lots of others

The __flag variable is turned into _Base__flag here. In Derived it is turned into _Derived__flag, so you are not actually overriding anything, just introducing a new variable.

If you want a variable to be overridable, do not use names staring with two underscores.

answered Oct 15, 2013 at 9:17

uranusjr

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