On Sun, Jan 16, 2022 at 04:50:40PM +0000, MRAB wrote:
> Not quite as bad as that:
>
> >>> f = frozenset({1, 2, 3})
> >>> f is frozenset(f)
> True
Mark suggested that on the bug tracker too, but that's not relevant. As
I replied there:
>>> 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.
--
Steve
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