All I really need to do is click [Check Syntax] button in DrRacket and all the modules/files in the program recompile. So there really is no problem. Took me a while to realize this. I would still like confirmation that it should be okay for me to clean up by removing all versions of Racket that are no longer current (by: raco pkg remove ...) And I would still like to know that if I don't remove previously installed versions of Racket, that I can switch between them. How does one make a previously installed version - current. (I might choose to keep them installed if it is possible to ocassionally make use of them. If it is not possible to make any of them current, then there is no point in keeping them on my system.) Thanks. Don.
On Tuesday, April 13, 2021 at 8:42:18 PM UTC-6 Philip McGrath wrote: > On Tue, Apr 13, 2021 at 10:31 PM John Clements <clem...@brinckerhoff.org> > wrote: > >> What if he has directories that aren’t part of an installed package? That >> was my concern, and why I suggested manually deleting compiled subdirs. >> > > That's right, but given: > > On Tue, Apr 13, 2021 at 8:16 PM Don Green <infodeve...@gmail.com> wrote: > >> Welcome to Racket v8.0 [cs]. >> > (current-library-collection-paths) >> '(#<path:/home/don/.plt-scheme/4.2.1/collects> >> #<path:/home/don/.racket/8.0/collects> >> #<path:/usr/share/racket/collects/>) >> >> My code files all begin here: /home/don/.plt-scheme/4.2.1/collects/... >> > > It looks like the code files are all linked as collections directly (i.e. > not via the package system). I think `raco setup --clean && raco setup` > should take care of anything under those directory trees (recursively). > > -Philip > -- 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/26a00595-092e-4b23-b3e9-233432e5dd8en%40googlegroups.com.
