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
