Is it explicitly specified?

[mario]

On Feb 3, 12:35 pm, TeroV <te... at nowhere.invalid> wrote:
> Jorge Godoy wrote:
> > mario ruggier wrote:
>
> >> Is there any way to tell between whether a keyword arg has been explicitly
> >> specified (to the same value as the default for it) or not... For example:
>
> >> def func(key=None):
> >>     do something with key
>
> >> But the following two usages give same results:
>
> >> func()
> >> func(key=None)
>
> >> It may sometimes be useful to make use of the conceptual difference
> >> between these two cases, that is that in one case the user did not specify
> >> any key and in the other the user explicitly specified the key to be None.
>
> >> Is there any way to tell this difference from within the called function?
> >> And if so, would there be any strong reasons to not rely on this
> >> difference? Would it be maybe a little abusive of what a keyword arg
> >> actually is?
>
> > If you change the idiom you use to:
>
> >>>> def myfunc(**kwargs):
> > ...     something = kwargs.get('something', None)
> > ...     print something
> > ...
> >>>> myfunc()
> > None
> >>>> myfunc(something='Something')
> > Something
>
> > Then you can test if 'something' is in kwargs dict or not.  If it isn't,
> > then you used the default value.  If it is, then the user
> > supplied 'something' to you, no matter what its value is.
>
> Exactly, and if you use idiom func(*args, **kwargs) you can distinguish
> all the usage cases:
>
>  >>> def func(*args, **kwargs):
> ...  print(args, kwargs)
> ...
>  >>> func()
> () {}
>  >>> func(key='alabama')
> () {'key': 'alabama'}
>  >>> func('alabama')
> ('alabama',) {}

Nice... but I would still like to be able to specify the key's default
value in the func signature, and in this case this would not be
possible:

>>> def func(key=None, *args, **kwargs):
...   print(key, args, kwargs)
...
>>> func()
(None, (), {})
>>> func(None)
(None, (), {})
>>> func(key=None)
(None, (), {})
>>>

I would still need an additional object, as Arnaud suggests.

mario