Static variable vs Class variable

[Marc 'BlackJack' Rintsch]

On Wed, 17 Oct 2007 00:33:59 -0700, paul.melis wrote:

> 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?

No.  As shown several times in this thread already.

> 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.

`__iadd__` *may* doing the addition in place and return `self` but it is
also allowed to return a different object.  So there is always a rebinding.

> 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']

Here you see that the method actually returns an object!

>>>> 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.

Because there is always a rebinding involved.

Ciao,
	Marc 'BlackJack' Rintsch