If you read the language reference for augmented arithmetic assignment, you
will note that it essentially says, "call __i__, and if that doesn't
work call as if you were doing a b". Unfortunately it appears **= does
not follow the rule of falling back on the binary arithmetic expression
semantics.
I am also for #2, as I don't think there is any concrete reason for making
**= special.
On Sun, Aug 23, 2020 at 5:05 PM Brett Cannon wrote:
> If you read the language reference for augmented arithmetic assignment,
> you will note that it essentially says, "call __i__, and if that
> doesn't work
On Sun, Aug 23, 2020 at 1:09 PM Brett Cannon wrote:
> If you read the language reference for augmented arithmetic assignment,
> you will note that it essentially says, "call __i__, and if that
> doesn't work call as if you were doing a b". Unfortunately it appears
> **= does not follow the rule
