Re: [Caml-list] question about modules

[Markus Mottl <mottl@miss.wu-wien.ac.at>]

On Wed, 06 Jun 2001, Mark Wotton wrote:
> I'm using the translation of Chris Okasaki's code that Markus Mottl has
> provided, and I've got a question about the way modules work. I had
> assumed that it worked in a similar way to lists: ie, i write "int list",
> i'd assumed I'd do something similar with Deques, "int Deque" for
> instance. Obviously this doesn't work: I understand that once I've added
> an element to a Deque, the polymorphic type is fixed, so there's no
> worries about type safety there; however, I had something like this:

I assume you have created a module "Deque" using one of the functors in
chapter 8, e.g.:

  module Deque = RealTimeDeque (struct let c = 3 end)

As you can see from the signature restrictions used in the functor head of
"RealTimeDeque", this functor generates modules that match the signature
"DEQUE":

  module RealTimeDeque (C : sig val c : int end) : DEQUE

Therefore, we have to look at this signature to see, what options we
have to use deques:

---------------------------------------------------------------------------
module type DEQUE = sig
  type 'a queue

  val empty : 'a queue
  val is_empty : 'a queue -> bool

  (* insert, inspect, and remove the front element *)
  val cons : 'a -> 'a queue -> 'a queue
  val head : 'a queue -> 'a        (* raises Empty if queue is empty *)
  val tail : 'a queue -> 'a queue  (* raises Empty if queue is empty *)

  (* insert, inspect, and remove the rear element *)
  val snoc : 'a queue -> 'a -> 'a queue
  val last : 'a queue -> 'a        (* raises Empty if queue is empty *)
  val init : 'a queue -> 'a queue  (* raises Empty if queue is empty *)
end
---------------------------------------------------------------------------

Obviously, the type used to refer to deques must be:

  'a Deque.queue

> type tree = CompTree of tree list * tree list;;
> 
> before I realised that I needed deques. It would seem that the translation
> is to
> 
> type tree = CompTree of Deque * Deque;;

Thus, you'll have to write:

  type tree = CompTree of tree Deque.queue * tree Deque.queue

If it bothers you to write so much, you can also define a type synonym
for deques:

  type 'a deque = 'a Deque.queue

and use it instead:

  type tree = CompTree of tree deque * tree deque

This is probably the most intuitive version.

Regards,
Markus Mottl

-- 
Markus Mottl, mottl@miss.wu-wien.ac.at, http://miss.wu-wien.ac.at/~mottl
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