Hi Andy, Andy C <andy.coolw...@gmail.com> wrote: > > >> user> (= s1 s2) >> true >> user> (= (seq s1) (seq s2)) >> false > > > Thx. If a=b then f(a) must = f(b). Something is broken here.
If a seq is a sequential view of a thing, and a set is an unordered thing, then it does not seem shocking to me that multiple sequential views of a given set, with different orderings, are possible. This may not be the only way to do things; and it may not be the way other languages do it; and it may not match your preference. But I think it's clearly wrong to say that it's internally inconsistent or "broken". It's perhaps hard to say this without sounding condescending, but rather than seeking to identify all the ways in which Clojure isn't Haskell, it might be more useful to pursue an understanding of Clojure (including its definitely-not-nonexistent flaws!) on its own terms. - John -- You received this message because you are subscribed to the Google Groups "Clojure" group. To post to this group, send email to clojure@googlegroups.com Note that posts from new members are moderated - please be patient with your first post. To unsubscribe from this group, send email to clojure+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/clojure?hl=en --- You received this message because you are subscribed to the Google Groups "Clojure" group. To unsubscribe from this group and stop receiving emails from it, send an email to clojure+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/groups/opt_out.
