# [plt-scheme] Predicates from Types

What if X and p don't actually match?
On Thu, Apr 2, 2009 at 9:39 AM, Paulo J. Matos <pocmatos at gmail.com> wrote:
>* On Thu, Apr 2, 2009 at 2:18 PM, Carl Eastlund <carl.eastlund at gmail.com> wrote:
*>>* On Thu, Apr 2, 2009 at 9:42 AM, Paulo J. Matos <pocmatos at gmail.com> wrote:
*>>>* On Thu, Apr 2, 2009 at 1:11 PM, Sam TH <samth at ccs.neu.edu> wrote:
*>>>>*
*>>>>* This isn't related to type inference, so that's not correct. But it
*>>>>* doesn't change the behavior of the function (no type annotation ever
*>>>>* changes the behavior of a Typed Scheme function).
*>>>*
*>>>* So, I guess I didn't really get it. What is the annotation that a
*>>>* function is a predicate for a type useful for?
*>>*
*>>* Let's say you have some type X, and some predicate p of type (Any ->
*>>* Boolean : X). Then you can write:
*>>*
*>>* (: f ((U X Y) -> X))
*>>* (define (f x-or-y)
*>>* (if (p x-or-y) x-or-y (error 'f "I don't have an X")))
*>>*
*>>* Notice that x-or-y initially has the type (U X Y), but by the time f
*>>* returns it in the first branch of the if it has type X. That's
*>>* because (p x-or-y) returns true, and the ": X" annotation tells Typed
*>>* Scheme that when p returns true, its input is of type X. So that ":
*>>* X" is useful when you have unions involving X and conditionals that
*>>* distinguish them.
*>>*
*>*
*>* Ahhhhhh, beautiful!!!
*>*
*>* Thanks for the explanation. In fact, going also back to what Sam said
*>* before about the number?, now I got it!
*>*
*>>* --
*>>* Carl Eastlund
*>>*
*>*
*>*
*>*
*>* --
*>* Paulo Jorge Matos - pocmatos at gmail.com
*>* Webpage: http://www.personal.soton.ac.uk/pocm
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