Re: [Caml-list] First Class modules -A bug in 4.00?

[bob zhang <bobzhang1988@gmail.com>]

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Hi, List,
  This works around the syntax error problem, but still does not work
------------------------------------------------------
module type S1 = sig
  module Inner  : sig
    type 'a t
    val add : 'a t -> 'a t -> int
  end
  type int_u = int Inner.t
end

let f1
    (type s)
    (type a)
    (module Y : S1  with type int_u = a) (y: Y.int_u) =
  Y.Inner.add y y

Jun 8, 2012 at 11:25 AM, bob zhang <bobzhang1988@gmail.com> wrote:

> Thanks. it helps.
> Here's what I really want, is there any way to walk around if not possible?
>
> S1 is not written by me.
> --------------------------------------------------------------------
> module type S1 = sig
>   module Inner  : sig
>     type 'a t
>     val add : 'a t -> 'a t -> int
>   end
> end
>
> let f1
>     (type s)
>     (type a)
>     (module Y : S1 with type s Inner.t = a) (y: s Y.Inner.t) =
>   Y.Inner.add y y
>
> ------------------------------------------------------------------------------------------------------------
> On Thu, Jun 7, 2012 at 11:22 PM, Jacques Garrigue <
> garrigue@math.nagoya-u.ac.jp> wrote:
>
>> On 2012/06/08, at 11:35, Hongbo Zhang wrote:
>>
>> > Hi, List
>> >  I am not sure this is a bug or not?
>> >
>> >  Below is a contrived example:
>> > ------------------
>> >  module type S = sig
>> >    type t=int
>> >    module  X : sig type u   end
>> >  end
>> >
>> >  let f ( module X : S)  (y:X.X.u) =
>> >  3
>> > --------------------
>> > Error: This pattern matches values of type X.X.u
>> >      but a pattern was expected which matches values of type X.X.u
>> >      The type constructor X.X.u would escape its scope
>> > -- Thanks
>>
>> Definitely, this is not a bug.
>> Type X.X.u is abstract, and showing it outside (as by taking an argument
>> of that type) would be meaningless.
>>
>> What you might have meant is:
>>
>>  let f (type a) (module X : S with type X.u = a) (y : X.X.u) = 3
>>
>> This is typable (but I'm not sure it means anything...)
>>
>> Jacques Garrigue
>>
>
>
>
> --
> -- Bob
>



-- 
-- Bob

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