Static variable vs Class variable

[paul.melis at gmail.com]

On Oct 10, 8:23 am, "Diez B. Roggisch" <de... at nospam.web.de> wrote:
> > However, it is not true that += "always leads to a rebinding of a to the
> > result of the operation +". The + operator for lists creates a new list.
> > += for lists does an in-place modification:
>
> It still is true.
>
> a += b
>
> rebinds a. Period. Which is the _essential_ thing in my post, because
> this rebinding semantics are what confused the OP.

Doesn't this depend on wether "a" supports __iadd__ or not? Section
3.4.7 of the docs say

"""
If a specific method is not defined, the augmented operation falls
back to the normal methods. For instance, to evaluate the expression x
+=y, where x is an instance of a class that has an __iadd__() method,
x.__iadd__(y) is called. If x is an instance of a class that does not
define a __iadd__() method, x.__add__(y) and y.__radd__(x) are
considered, as with the evaluation of x+y.
"""

So if a.__iadd__ exists, a += b is executed as a.__iadd__(b), in which
case there's no reason to rebind a.

However, this confuses the heck out of me:

>>> class A:
...     l = []
...
>>> class B(A): pass
...
>>> B.__dict__
{'__module__': '__main__', '__doc__': None}
>>> B.l
[]
>>> B.l.append('1')
>>> B.l
['1']
>>> B.__dict__
{'__module__': '__main__', '__doc__': None}
>>> B.l.__iadd__('2')
['1', '2']
>>> B.l
['1', '2']
>>> B.__dict__
{'__module__': '__main__', '__doc__': None}
>>> B.l += '3'
>>> B.__dict__
{'__module__': '__main__', '__doc__': None, 'l': ['1', '2', '3']}

Why is B.l set for the += case only? B.l.__iadd__ obviously exists.

Paul