[Python-ideas] Keyword only argument on function call

[Steven D'Aprano]

On Wed, Sep 12, 2018 at 02:23:34PM +0100, Jonathan Fine wrote:
> Steve Barnes suggested adding __params__, as in
> 
> >      def a_method(self, cr, uid, ids, values, context=None):
> >        ...
> >        params = {k:v for k,v in __params__ if k in parent.a_method.keys()}
> >        # Possibly add some additional entries here!
> >        super(self, parent).a_method(**params)

In context, what Steve Barnes said was

    If this [__params__] was accessible externally, 
    as fn.__defaults__ is [...]

https://mail.python.org/pipermail/python-ideas/2018-September/053322.html

Here is the behaviour of fn.__defaults__:

py> def fn(a=1, b=2, c=3):
...     pass
...
py> fn.__defaults__
(1, 2, 3)


Notice that it is an externally acessible attribute of the function 
object. If that's not what Steve Barnes meant, then I have no idea why 
fn.__defaults__ is relevant or what he meant.

I'll confess that I couldn't work out what Steve's code snippet was 
supposed to mean:

    params = {k:v for k,v in __params__ if k in parent.a_method.keys()}

Does __params__ refer to the currently executing a_method, or the 
superclass method being called later on in the line?

Why doesn't parent.a_method have parens?

Since parent.a_method probably isn't a dict, why are we calling keys() 
on a method object?

The whole snippet was too hard for me to comprehend, so I went by the 
plain meaning of the words he used to describe the desired semantics.

If __params__ is like fn.__defaults__, then that would require setting 
fn.__params__ on each call.

Perhaps I'm reading too much into the "accessible externally" part, 
since Steve's example doesn't seem to actually be accessing it 
externally.



-- 
Steve