This code is looking better. However, [EqSet] retains the same problem:
its output signature hides the equality [a = Q.item]. You really want
the output signature to contain just a [type t] with no [a]. Then just
use [Q.item] in place of [a] and [t] in place of [t a], everywhere in
the signature.
P.S.: You can omit the line for [eq_item] at the very end, and the
compiler will infer it, just as you'd expect with type classes!
On 02/10/2015 07:37 AM, Gabriel Riba wrote:
Thanks for the hint.
type t a = list Q.item
(* --- new code: *)
functor EqSet(Q: sig con item :: Type
val eq_item: eq item
end): sig
con t :: Type -> Type
con a :: Type
val empty: t a
val singleton : a -> t a
val insert: a -> t a -> t a
val member: a -> t a -> bool
val foldr: b ::: Type -> (a -> b -> b) -> b -> t a -> b
end = struct
type a = Q.item
type t a = list Q.item
val empty: t a = []
val singleton (x: a): t a = x :: []
val member (x: a) (li: t a) = List.exists (eq x) li
val insert (x: a) (xs : t a) =
if member x xs then xs else x :: xs
val foldr [b] (myop: a -> b -> b) (z: b) (li: t a) = List.foldr myop z li
end
structure IntEqSet = EqSet(struct type item = int
val eq_item = eq_int
end)
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