Sam Tobin-Hochstadt writes: > I recommend doing the mutation yourself, and not using `shared`. > That's the most obvious solution.
Not for me, unfortunately, but probably I am missing something obvious. Here's explicit mutation in untyped Racket: #lang racket (struct foo (x) #:mutable) (struct bar (x)) (define f (foo (void))) (define b (bar f)) (set-foo-x! f b) (eq? (bar-x b) f) (eq? (foo-x f) b) That works fine. Moving to Typed Racket, this is rejected by the type checker because (void) is of type Void. Since I need a bar to make a foo and a foo to make a bar, I don't see how I can ever initialize my foo structure. > The reason that your `require/typed` doesn't work is that it creates > new versions of the placeholder functions, but those aren't the ones > that `shared` uses internally, so Typed Racket doesn't use the types > you've given. That makes sense, thanks for the explanation! Konrad. ____________________ Racket Users list: http://lists.racket-lang.org/users
