On Mon, 17 Jan 2022 at 01:11, Chris Angelico <ros...@gmail.com> wrote:
>
> On Mon, Jan 17, 2022 at 10:14 AM Paul Moore <p.f.mo...@gmail.com> wrote:
> >
> > On Sun, 16 Jan 2022 at 22:55, Steven D'Aprano <st...@pearwood.info> wrote:
> > > >>> def f():
> > > ... return frozenset({1, 2, 3})
> > > ...
> > > >>> a = f.__code__.co_consts[1]
> > > >>> a
> > > frozenset({1, 2, 3})
> > > >>> b = f()
> > > >>> assert a == b
> > > >>> a is b
> > > False
> > >
> > > Each time you call the function, you get a distinct frozenset object.
> >
> > I may just be reiterating your point here (if I am, I'm sorry - I'm
> > not completely sure), but isn't that required by the definition of the
> > frozenset function. You're calling frozenset(), which is defined to
> > "Return a new frozenset object, optionally with elements taken from
> > iterable". The iterable is the (non-frozen) set {1, 2, 3}.
>
> Where is that definition?
https://docs.python.org/3/library/functions.html#func-frozenset
> According to help(frozenset), it will
> "[b]uild an immutable unordered collection of unique elements", so it
> doesn't necessarily have to be a brand new object.
I'd read "build" as meaning it is a new object. But regardless, I view
the language documentation as authoritative here.
> With mutables, it
> does have to return a new one every time (list(x) will give you a
> shallow copy of x even if it's already a list), but with immutables,
> it's okay to return the same one (str and tuple will return self).
That's an optimisation, not a guarantee, so while it's relevant, it's
not sufficient (IMO).
However, it's not clear how relevant this is to the actual proposal,
so the precise details probably aren't that important.
Paul
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