On Oct 19, 2007, at 01.19h, Alan Falloon wrote:

> I am having a really hard time figuring out how to instantiate a  
> functor when its arguments have module constraints. Here is an  
> example:
>
> module type PIE = sig
>   type pie
>   val bake : unit -> pie
>   val eat : pie -> unit
> end
>
> module type BAKER = sig
>   module Pie : PIE
>   val dozen : unit -> Pie.pie array
> end
>
> module type PIG = sig
>   module Pie : PIE
>   val feed : Pie.pie -> unit
> end
>
> (* Need a module constraint to make sure that the baker and pig are  
> talking about the same pie *)
> module Farmer (Baker : BAKER) (Pig : PIG with module Pie =  
> Baker.Pie) = struct
>   let feed_pigs () =
>     let pies = Baker.dozen () in
>       Array.iter Pig.feed pies
> end
>
> module Apple : PIE = struct
>   type pie = Pie of string
>   let bake () = Pie "apple"
>   let eat (Pie "apple") = ()
> end
>
>
> module Joe = Farmer(Bob)(Daisy) (*BOOM*)
> let () = Joe.feed_pigs()
>
>
> However, this fails to compile because when I try to make module  
> Joe because Bob.Pie and Daisy.Pie aren't the same. Fair enough,  
> thats what abstraction is for, however I can't figure out the  
> syntax to instantiate Joe! I tried this:
>
> module Joe = Farmer(Bob)(Daisy:PIG with module Pie=Bob.Pie)
>

As you noted, abstraction prevents the types in the Pie submodules  
from being equivalent. And there is no way to retract that  
abstraction later - it wouldn't be abstraction otherwise. So you will  
not be able to apply Joe as you tried. This is not a question of syntax.

The solution is to avoid the over-abstraction and keep the Pie  
submodules transparent. Try:


> module Bob : (BAKER with module Pie = Apple) = struct
>   module Pie = Apple
>   let dozen () = Array.init 13 (fun _ -> Pie.bake ())
> end
>
> module Daisy : (PIG with module Pie = Apple) = struct
>   module Pie = Apple
>   let feed p = Pie.eat p; print_endline "OINK!"
> end


- Andreas