[racket] [Typed Racket] define-predicate for union with functions

You can sort of fake it if you wrap your function in a structure type.  In
lieu of

(define-type T1 (Symbol -> String))

Instead use a struct:.

(struct: T1 ([f : (Symbol -> String)]))

;; predicate is T1? for a wrapped Symbol->String function

(define-type T2 (U Symbol Number T1)) ;; T1 is a boxed function of
Symbol->String

(: yup (T2 -> String))
(define (yup t2)
  (cond
    ((symbol? t2) (symbol->string t2))
    ((number? t2) (number->string t2))
    (else (match t2
            ((T1 f) (f 'goodbye))))))

(yup 'hello)
(yup 123)
(yup (T1 symbol->string))

And I don't know but maybe it would be possible for a sufficiently smart
compiler to elide the T1 boxing and unboxing of the wrapped function.
Well for that matter eliding any single wrapped value type.


On Tue, May 21, 2013 at 9:54 AM, Matthias Felleisen <matthias at ccs.neu.edu>wrote:

>
> (And this is not a limitation of Typed Racket but a fundamental problem of
> CS.)
>
>
> On May 20, 2013, at 11:07 AM, Eric Dobson <eric.n.dobson at gmail.com> wrote:
>
> > No, you cannot define predicates for function types. If you explain
> > the problem you have, we might be able to explain another solution
> > that will work.
> >
> > On Mon, May 20, 2013 at 1:32 AM, Paul Leger <pleger at gmail.com> wrote:
> >> Hi all,
> >>      Maybe, this question is very easy. In the following code, I try
> >> defining a predicate for T2, but I cannot because T1 it is a function.
> >>
> >> (define-type T1 (Symbol -> Any) )
> >> (define-type T2 (U Symbol Number T1))
> >>
> >> ;(define-predicate T1? Procedure) ;this line is useless
> >>
> >> (define-predicate T2? T2)
> >>
> >>> Type Checker: Type T2 could not be converted to a contract. in: T2
> >>
> >> My unsuccessful answer is:
> >>  (define-type T2 (U Symbol Number Procedure))
> >>
> >> I do not like it because I lose the relationship between T1 and T2. Are
> >> there some possibility
> >>
> >> Thank in advance,
> >> Paul
> >>
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