On 2014-05-10 11:30:58 -0400, Alexander D. Knauth wrote: > After trying some stuff and guessing stuff from error messages, and just > to try it and see what happened, I tried this: > > (: string-with-length-1? (Any -> Boolean : #:+ (String @ x) #:- ((U String > Any) @ x))) > (define (string-with-length-1? x) > (if (string? x) > (one? (string-length x)) > #f)) > > And it gave me this weird error message: > . Type Checker: type mismatch; > mismatch in filter > expected: ((String @ x) | Top) > given: ((String @ x) | Top) in: (if (string? x) (one? (string-length x)) > #f) > What? If it*s expecting ((String @ x) | Top), and I*m giving it ((String > @ x) | Top), then what*s the problem?
This works for me if I use the following type declaration instead: (: string-with-length-1? (Any -> Boolean : #:+ (String @ 0) #:- ((U String Any) @ 0))) Note that I've switched `x` with `0` in the filters. Basically, the formal parameter `x` is not in scope when you write a type declaration like this *outside* of the function. If you have a type inside the function, it should be in scope. The `0` means that you're referring to the 1st parameter, which is `x`. > And is there any documentation anywhere about this? There is now, but I just added it a few days ago and it's only in the pre-release docs (the syntax may change by v6.0.2): http://www.cs.utah.edu/plt/snapshots/current/doc/ts-reference/type-ref.html?q=#%28form._%28%28lib._typed-racket%2Fbase-env%2Fbase-types-extra..rkt%29._-~3e%29%29 In the pre-release, you can also write this instead: (: string-with-length-1? (Any -> Boolean : #:+ String)) which is a shorthand for the type above. It'll also print like that. Cheers, Asumu ____________________ Racket Users list: http://lists.racket-lang.org/users
