python - Class that acts as mapping for **unpacking

Question

Without subclassing dict, what would a class need to be considered a mapping so that it can be passed to a method with **.

from abc import ABCMeta

class uobj:
    __metaclass__ = ABCMeta

uobj.register(dict)

def f(**k): return k

o = uobj()
f(**o)

# outputs: f() argument after ** must be a mapping, not uobj

At least to the point where it throws errors of missing functionality of mapping, so I can begin implementing.

I reviewed emulating container types but simply defining magic methods has no effect, and using ABCMeta to override and register it as a dict validates assertions as subclass, but fails isinstance(o, dict). Ideally, I dont even want to use ABCMeta.

Answer 1

The __getitem__() and keys() methods will suffice:

>>> class D:
        def keys(self):
            return ['a', 'b']
        def __getitem__(self, key):
            return key.upper()


>>> def f(**kwds):
        print kwds


>>> f(**D())
{'a': 'A', 'b': 'B'}