frozenset can be altered by |=

[Chris Angelico]

On Tue, Nov 30, 2021 at 12:41 PM Richard Damon <Richard at damon-family.org> wrote:
>
> On 11/29/21 5:01 PM, Chris Angelico wrote:
> > On Tue, Nov 30, 2021 at 8:55 AM Marco Sulla
> > <Marco.Sulla.Python at gmail.com> wrote:
> >> I must say that I'm reading the documentation now, and it's a bit
> >> confusing. In the docs, inplace operators as |= should not work. They
> >> are listed under the set-only functions and operators. But, as we saw,
> >> this is not completely true: they work but they don't mutate the
> >> original object. The same for += and *= that are listed under `list`
> >> only.
> >>
> > Previously explained here:
> >
> >>> On Mon, 22 Nov 2021 at 14:59, Chris Angelico <rosuav at gmail.com> wrote:
> >>>> Yeah, it's a little confusing, but at the language level, something
> >>>> that doesn't support |= will implicitly support it using the expanded
> >>>> version:
> >>>>
> >>>> a |= b
> >>>> a = a | b
> >>>>
> >>>> and in the section above, you can see that frozensets DO support the
> >>>> Or operator.
> >>>>
> >>>> By not having specific behaviour on the |= operator, frozensets
> >>>> implicitly fall back on this default.
> >>>>
> > The docs explicitly show that inplace operators are defined for the
> > mutable set, and not defined for the immutable frozenset. Therefore,
> > using an inplace operator on a frozenset uses the standard language
> > behavior of using the binary operator, then reassigning back to the
> > left operand.
> >
> > This is a language feature and applies to everything. You've seen it
> > plenty of times with integers, and probably strings too. A frozenset
> > behaves the same way that anything else does.
> >
> > ChrisA
>
> I suppose the question comes down to is it worth adding a reminder in
> the description of the inplace operators that if a type doesn't support
> the inplace operator, it is automatically converted into the equivalent
> assignment with the binary operator?
>

My view is: no, because you'd have to put that reminder on every
single object in Python. The details are here, and apply to all Python
code, not to any particular type:

https://docs.python.org/3/reference/simple_stmts.html#augmented-assignment-statements

ChrisA