No, I don't think that's it. The issue is that one is a mutable set and the
other isn't, so they aren't equal (even if their elements aren't equal).

> (equal? (mutable-seteqv) (list->seteqv '()))
#f

Maybe you wanted to call list->mutable-seteqv? Or maybe just start with an
immutable set?

Robby


On Tue, Dec 8, 2020 at 6:40 PM Nathaniel W Griswold <[email protected]>
wrote:

> Thanks. Switching to 7.9 now.
>
> Nate
>
> > On Dec 8, 2020, at 6:38 PM, Jon Zeppieri <[email protected]> wrote:
> >
> > I think that's this bug
> > [
> https://github.com/racket/racket/commit/543dab59640fa5e911443baaadaae471406dbf40
> ],
> > which should be fixed in 7.9. - Jon
> >
> > On Tue, Dec 8, 2020 at 7:19 PM Nathaniel W Griswold
> > <[email protected]> wrote:
> >>
> >> I don’t know if i’m missing something or what, but the following is
> confusing me:
> >>
> >> (let ([test (mutable-seteqv)])
> >>  (for* ([i (in-range 1000)]
> >>         [j (random 0 1000)])
> >>    (set-add! test j))
> >>  (let ([test-copy (set-copy test)])
> >>    (printf "test-copy=~a\n" (set->list test-copy))
> >>    (printf "Equal from stream is ~a\n" (equal? (list->seteqv (set->list
> test-copy)) test))))
> >>
> >> prints something like:
> >> test-copy=(31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12
> 11 10 9 8 7 6 5 4 3 2 1 0)
> >> Equal from stream is #f
> >> Equal regular is #t
> >>
> >>
> >> Why is this?
> >>
> >> Thanks
> >>
> >> Nate
> >>
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