Looking for magic method to override to prevent dict(d) from grabbing subclass inst d contents directly

[Bengt Richter]

On Tue, 22 Nov 2005 09:30:49 -0500, Mike Meyer <mwm at mired.org> wrote:

>bokr at oz.net (Bengt Richter) writes:
>> Has anyone found a way besides not deriving from dict?
>> Shouldn't there be a way?
>> TIA
>> (need this for what I hope is an improvement on the Larosa/Foord OrderedDict ;-)
>>
>> I guess I can just document that you have to spell it dict(d.items()), but I'd
>> like to hide the internal shenanigans ;-)
>
>So why not just hide them:
>
>>>> class md(dict):
>...  def __new__(cls, arg):
>...   return dict.__new__(cls, arg.items())
>
>
>I'm not sure exactly what you mean by "grabbing sublass inst d contens
>directly", but if dict(d.items()) does it, the above class should do
>it as well. Of course, *other* ways of initializing a dict won't work
>for this class.
>
It's not initializing the dict subclass that is a problem, it is passing
it to the plain dict builtin as a constructor argument and being able to
control what is pulled from the subclass object to do the construction
of the ordinary dict.

Here's where I am on this: (odictb is my version of odict ;-)

 >>> from odictb import OrderedDict
 >>> d = OrderedDict()
 >>> d['a'] = 1
 >>> d['b'] = 2
 >>> d
 {'a': 1, 'b': 2}
 >>> dict(d.items())
 {'a': 1, 'b': 2}

Looks normal, but I am secretly using key:(insertionseqno, value) to represent key:value

  >>> dict(d)
 {'a': (0, 1), 'b': (1, 2)}

and I want to control dict(d) that so the result is like dict(d.items()) above.

The dict builtin apparently skips looking for methods in the dict subclass. If it were
a subclass of object, dict would probably be looking for __iter__ or __getitem__ or __???___
but with a dict subclass it seems to bypass such a check. I was hoping that even so
there might be a __???__ that I didn't know about.

[Oop, just discovered a bug in my implementation ;-/ Hope it doesn't ripple...]

Regards,
Bengt Richter