On 11/29/21 5:01 PM, Chris Angelico wrote:
On Tue, Nov 30, 2021 at 8:55 AM Marco Sulla
<marco.sulla.pyt...@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 <ros...@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
--
https://mail.python.org/mailman/listinfo/python-list
