Learning Descriptors
Ian Kelly
ian.g.kelly at gmail.com
Sun Jul 31 09:03:39 EDT 2016
On Sun, Jul 31, 2016 at 6:33 AM, Gerald Britton
<gerald.britton at gmail.com> wrote:
> Today, I was reading RH's Descriptor HowTo Guide at
>
> https://docs.python.org/3/howto/descriptor.html?highlight=descriptors
>
> I just really want to fully "get" this.
>
> So I put together a little test from scratch. Looks like this:
>
> class The:
> class Answer:
> def __get__(self, obj, type=None):
> return 42
>
>>>> The.Answer
> <class '__main__.The.Answer'>
>>>>
>
> but, I expected to see 42.
>
> So, digging deeper I read:
>
> For classes, the machinery is in type.__getattribute__() which transforms
> B.x into B.__dict__['x'].__get__(None, B). In pure Python, it looks like:
>
> def __getattribute__(self, key):
> "Emulate type_getattro() in Objects/typeobject.c"
> v = object.__getattribute__(self, key)
> if hasattr(v, '__get__'):
> return v.__get__(None, self)
> return v
>
> OK, so I copied this function, then ran it and got:
>
>>>> __getattribute__(The, 'Answer')
> 42
>
> So, what I don't get is why the "B.x into B.__dict__['x'].__get__(None, B)"
> part doesn't work in my case.
>
> I'm sure I'm missing something here (that`s usually the case for me <:‑|) ,
> but what?
Obviously that __getattribute__ is not exactly like the real one. The
real one looks up __get__ as a *method* of B.__dict__['x'], which
requires that B.__dict__['x'] be an instance of some class that
defines __get__, not the class itself. Try this:
py> class Answer:
... def __get__(self, obj, type):
... return 42
...
py> class The:
... Answer = Answer()
...
py> The.Answer
42
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