Re: [Caml-list] Private types

[Jacques Garrigue <garrigue@math.nagoya-u.ac.jp>]

Dear Dario,

Since you use a private abbreviation, extraction from the private type
must be explicit before you can do anything on its representation.
So the solution is simple enough:

module Mymod =
  struct
    let is_foo2 x =
      match (x : _ Foobar.t :> [> ]) with `A -> true | `B -> false
  end

The [> ] as target type just says that you expect a polymorphic
variant, which is incompatible with Foobar.t, and triggers its
expansion.

If you want to avoid this explicit coercion, you must use a private
row type in place of a private abbreviation. However, you loose the
ability to distinguish between values created by make_a and make_b at
the type level.

module Foobar: sig
  type foo_t = [ `A ]
  type bar_t = [ `B ]
  type foobar_t = [ foo_t | bar_t ]
  type t = private [< foobar_t]

  val make_a: unit -> t
  val make_b: unit -> t
  val is_foo: t -> bool
  val is_bar: t -> bool
end = struct
  type foo_t = [ `A ]
  type bar_t = [ `B ]
  type foobar_t = [ foo_t | bar_t ]
  type t = foobar_t

  let make_a () = `A
  let make_b () = `B
  let is_foo = function `A -> true | `B -> false
  let is_bar = function `B -> true | `A -> false
end

You may recover this distinction by adding an extra parameter to t,
just as in your original code.

  type 'a t = private [< foobar_t]

but since you won't be able to relate it directly to the expansion of
the type (the row variable is quantified at the module level, so you
cannot capture it with 'a), you would have to recover it by hand.

Jacques Garrigue

> I have also been playing with 3.11's private types, and I would like to
> share a problem I've come across.  Suppose I have a Foobar module defined
> as follows (note the use of a type constraint):
> 
> 
> module Foobar:
> sig
> 	type foo_t = [ `A ]
> 	type bar_t = [ `B ]
> 	type foobar_t = [ foo_t | bar_t ]
> 	type 'a t = 'a constraint 'a = [< foobar_t ]
> 
> 	val make_a: unit -> foo_t t
> 	val make_b: unit -> bar_t t
> 	val is_foo: [< foobar_t] t -> bool
> 	val is_bar: [< foobar_t] t -> bool
> end =
> struct
> 	type foo_t = [ `A ]
> 	type bar_t = [ `B ]
> 	type foobar_t = [ foo_t | bar_t ]
> 	type 'a t = 'a constraint 'a = [< foobar_t ]
> 
> 	let make_a () = `A
> 	let make_b () = `B
> 	let is_foo = function `A -> true | `B -> false
> 	let is_bar = function `B -> true | `A -> false
> end
> 
> 
> Suppose also that I want to define a "is_foo" function in an external module.
> This function only needs to pattern-match on Foobar.t values.  The following
> code will work:
> 
> module Mymod:
> sig
> 	open Foobar
> 
> 	val is_foo2: [< foobar_t] t -> bool
> end =
> struct
> 	let is_foo2 = function `A -> true | `B -> false
> end
> 
> 
> But now consider that I want to enforce the creation of Foobar.t values only
> via Foobar's constructor functions, but I would like to keep the possibility of
> external modules to pattern-match on Foobar.t values.  In other words, change
> Foobar but don't break Mymod.  The immediate (naïve) solution is to make
> use of private types, thus changing the signature of Foobar to the following:
> 
> 
> module Foobar:
> sig
> 	type foo_t = [ `A ]
> 	type bar_t = [ `B ]
> 	type foobar_t = [ foo_t | bar_t ]
> 	type 'a t = private 'a constraint 'a = [< foobar_t ]
> 
> 	val make_a: unit -> foo_t t
> 	val make_b: unit -> bar_t t
> 	val is_foo: [< foobar_t] t -> bool
> 	val is_bar: [< foobar_t] t -> bool
> end = (...)
> 
> 
> But this will break Mymod.  The compile will complain with the following error:
> 
>       Values do not match:
>          val is_foo2 : [< `A | `B ] -> bool
>        is not included in
>          val is_foo2 : [< Foobar.foobar_t ] Foobar.t -> bool
> 
> Any ideas on how can I get around this problem?
> 
> Thanks!
> Cheers,
> Dario Teixeira