Having thought about it some more, I agree with Daniel Moisset that the
current behavior is correct -- in D(__foo=bar), __foo is a keyword arg
position and those don't get mangled. Anyway, anything mangled would get
mangled according to the containing class (in your example, C), not
according to th
Guido van Rossum wrote:
> If you want any kind of traction on this I recommend filing an opinionated
> issue on this (explaining why the current behavior is wrong).
Thanks - I'm asking from the point of view of trying to reimplement it. I don't
actually have a strong opinion on whether the curren
If you want any kind of traction on this I recommend filing an opinionated
issue on this (explaining why the current behavior is wrong).
On Wed, Jun 15, 2022 at 1:25 PM wrote:
> Daniel Moisset wrote:
> > I might expect that in a "case D(something=__y)" you get the mangling for
> > __y, but I'm n
Daniel Moisset wrote:
> I might expect that in a "case D(something=__y)" you get the mangling for
> __y, but I'm not sure what the implementation does now and I'm writing from
> my phone
Yes - that case does what you'd expect.
Thanks for the reply.
___
I don't remember this topic ever being discussed, but still I wouldn't
expect that __something to be mangled, given that it refers to an attribute
of an instance of D.
I might expect that in a "case D(something=__y)" you get the mangling for
__y, but I'm not sure what the implementation does now a
