Why would an object always be linked to a variable name, and why only one
name?
Le mar. 12 sept. 2023 à 10:23, Dom Grigonis <dom.grigo...@gmail.com> a
écrit :
> As far as I understand, it is not possible. Is it?
>
> Something similar to:
>
> var = 710
> variable_name = [k for k, v in locals().items() if v == 710][0]
> print("Your variable name is " + variable_name)
>
> ,except robust.
>
> Possible ways to expose it:
> * builtin function `varname(obj: object) —> str`
> * object.__varname__ dunder (this one wouldn’t clutter general namespace)
>
> I am working on some code and just got to the point where it seems I could
> make use of it. The case is as follows:
>
> # What I do now:class A:
> defaults = dict()
>
> def set_default(self, a):
> self.defaults['a'] = a
>
> def make_something(self, a=None):
> a = a if a is not None else self.defaults.get('a')
> return Something(a)
> # What I would like to be able to do:class A:
> defaults = dict()
>
> def set_default(self, a):
> self.defaults[a.__varname__] = a
>
> def make_something(self, a=None):
> a = a if a is not None else self.defaults.get(a.__varname__)
> return Something(a)
>
>
> In this case I like second one, because it allows a tight coupling of
> variable names, which has the following benefits:
> 1. General simplification, where one doesn’t have to think variable name
> and its key value in dictionary. Code looks cleaner.
> 2. If one was to change a variable name of `set_default` it would
> immediately surface as an error.
> 3. For someone working with IDEs or making use of clever multiple
> selection tools, it would be easy to change all in one go.
>
> Any reasons why the above would not be a good practice?
> Any reasons why exposing it is not a good idea?
> Would this be difficult to achieve from python dev’s perspective?
>
>
> — This one can’t be solved with deferred eval :/ —
> Regards,
> DG
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--
Antoine Rozo
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