frozenset can be altered by |=

[Richard Damon]

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?

-- 
Richard Damon