(eq? +nan.0 -nan.0) -> #t This is documented: The datum -nan.0 refers to the same constant as +nan.0, and -nan.f is the same as +nan.f. Jos
_____ From: racket-users@googlegroups.com [mailto:racket-users@googlegroups.com] On Behalf Of David Storrs Sent: martes, 06 de febrero de 2018 00:42 To: Racket Users Subject: [racket-users] (number->string -nan.0) == "+nan.0" ? I noticed that (number->string -nan.0) yields "+nan.0" instead of "-nan.0" as I would have expected. It's not an issue for me, but I was wondering why this is? -- 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. For more options, visit https://groups.google.com/d/optout. -- 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. For more options, visit https://groups.google.com/d/optout.
