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.
On Mon, 22 Nov 2021 at 19:54, Marco Sulla <marco.sulla.pyt...@gmail.com> wrote: > > Yes, and you do this regularly. Indeed integers, for example, are immutables > and > > a = 0 > a += 1 > > is something you do dozens of times, and you simply don't think that > another object is created and substituted for the variable named `a`. > > On Mon, 22 Nov 2021 at 14:59, Chris Angelico <ros...@gmail.com> wrote: > > > > On Tue, Nov 23, 2021 at 12:52 AM David Raymond <david.raym...@tomtom.com> > > wrote: > > > It is a little confusing since the docs list this in a section that says > > > they don't apply to frozensets, and lists the two versions next to each > > > other as the same thing. > > > > > > https://docs.python.org/3.9/library/stdtypes.html#set-types-set-frozenset > > > > > > The following table lists operations available for set that do not apply > > > to immutable instances of frozenset: > > > > > > update(*others) > > > set |= other | ... > > > > > > Update the set, adding elements from all others. > > > > 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. > > > > ChrisA > > -- > > https://mail.python.org/mailman/listinfo/python-list -- https://mail.python.org/mailman/listinfo/python-list
