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Extending Set - strange behavior of abstract type
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Dawid Toton
- rossberg@m...
- Sylvain Le Gall
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Date: | 2010-04-27 (14:45) |
From: | Sylvain Le Gall <sylvain@l...> |
Subject: | Re: Extending Set - strange behavior of abstract type |
On 27-04-2010, Dawid Toton <d0@wp.pl> wrote: > I tried to extend the standard Set module with new operations. I got > error messages about type incompatibilities (the Set.S.t as exposed by > my implementation and Set.S.t used by functions from the original Set). > I have reduced my code to the following small example: > > module Set = struct > module Make (Ord : Set.OrderedType) = struct > module Set = Set.Make(Ord) > include Set > end > end > > module OrdChar = struct type t = char let compare = compare end > module Raw1 = Set.Make (OrdChar) > module Raw2 = Set.Make (struct type t = char let compare = compare end) > > let aaa (aa : Raw1.t) (bb : Raw1.Set.t) = (aa = bb) > let aaa (aa : Raw2.t) (bb : Raw2.Set.t) = (aa = bb) > > Only the last line results in an error: > Error: This expression has type Raw2.Set.t but is here used with type Raw2.t > > All the rest of the code compiles correctly. It means that types Raw1.t > and Raw1.Set.t can be unified. > > My question is: why these nearly identical statements results in > different behavior of the type t? > > I'd really prefer Raw1 and Raw2 to be identical. You just have to propagate the type by hand: module Set = struct module Make (Ord : Set.OrderedType) = struct include Set.Make(Ord) module Set : Set.S with type t = t = Set.Make(Ord) end end The "type t = t" do the trick. The first t is bound inside Set and the other comes from "include Set.Make(Ord)". Regards Sylvain Le Gall