On Thu, Feb 12, 2015 at 3:33 AM, Keean Schupke <[email protected]> wrote: > To me, (f x y) and ((f x) y) are the same thing, in both cases f :: a -> b > -> c where the types of a, b, and c are inferred.
But in Shap's Option 1, you have (fn 2 a -> b -> c) and (fn 1 a -> fn 1 b -> c) as distinct types which are not implicitly coerced to one another. So if you only see f used in (f x y), and you consider that to be the same as ((f x) y), which one is the type of f? _______________________________________________ bitc-dev mailing list [email protected] http://www.coyotos.org/mailman/listinfo/bitc-dev
