On Fri, Oct 25, 2019 at 9:28 AM Alexis King <lexi.lam...@gmail.com> wrote:
> Unlike eq? on symbols, eq?’s behavior on quoted lists is unspecified, so I > do not think there is a significantly deeper reason than “that isn’t what > the current implementation chooses to do.” Whether the answer is #t or #f > could change tomorrow, on a different VM, on a different architecture, or > on Friday the 13th. > > Is there a reason you would like the answer to be #t? Not strong one. I was implementing a compiler (to a computer simulator that I did) and I wanted to express some of my constants as list (because it would make easier to read them in case expression). I switched to use match instead. My intuition was that two quoted list of constant values would eq?. > > > On Oct 25, 2019, at 09:34, wanderley.guimar...@gmail.com wrote: > > > > Why (eq? (quote a) (quote a)) is #t but (eq? (quote (a)) (quote (a))) > > is #f? I would expect that if (quote (a)) was a mutable pair but it > > is not since (quote (a)) returns #f. It seems that guile returns #t > > as I was expecting. > > -- Abraço, Wanderley Guimarães -- You received this message because you are subscribed to the Google Groups "Racket Users" group. To unsubscribe from this group and stop receiving emails from it, send an email to racket-users+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/racket-users/CAAmHZoeGAQs%2BD0GzkVivzE559whPxnie19BDK4UnzR0gqWCyMQ%40mail.gmail.com.
