Question

Here’s a bit of a newbie Python question about instance variables.

Consider the following Python 2.7 class definition:

class Foo(object):
    a = 1

    def __init__(self):
            self.b = 2

    def __repr__(self):
            return "%s" % self.__dict__

Now, when I create an instance of Foo, Foo.__dict__ contains b, but not a.

>>> x=Foo()
>>> x
{'b': 2}
>>> dir(x)
['__class__', '__delattr__', '__dict__', '__doc__', '__format__', 
 '__getattribute__', '__hash__', '__init__', '__module__', '__new__', 
 '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__sizeof__',
 '__str__', '__subclasshook__', '__weakref__', 'a', 'b']
>>> x.__dict__
{'b': 2}

And here I thought I had a pretty good grasp on the Way of the Python.

What’s the difference between x.a and x.b? As far as I can tell they’re both instance variables.

Edit: OK, re-reading the Python docs I see that Foo.a is a class attribute rather than an instance variable. Hm… I guess the confusion comes from the fact that I can assign a new value to x.a and the new value only affects the x instance — I suppose I’m now aliasing a member variable over the top of the Foo.a attribute:

>>> y=Foo()
>>> y.a = 2
>>> y
{'a': 2, 'b': 2}
>>> x
{'b': 2}
>>> x.a
1
>>> z=Foo()
>>> z
{'b': 2}
>>> z.a
1
>>> Foo.a
1
>>> x.a
1
>>> y.a
2

So, now I overwrite the previous value of Foo.a, and it affects all instances of Foo that haven’t aliased Foo.a:

>>> Foo.a=999
>>> x.a
999
>>> y.a
2

Answer 1

Your a isn’t an instance variable. You defined it as part of the class.

>>> class Foo(object):
...    a = 1
...
>>> Foo.a
1

If you want an instance variable you should put it inside the __init__ method, because this method is called when your object is created.