Re: [Caml-list] Functor signature "inheritance"

[Nicolas Ojeda Bar <nicolas.ojeda.bar@lexifi.com>]

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Hi Shayne,

I did not try it, but I think you want

module Mul_default : MUL = functor (E : EQ) (N : NUM with type t = E.t)  ->
struct
    include E
    include (N : NUM with type t := E.t)

    let mul (x : t) (y : t) : t = failwith "foo"
end

Cheers,
Nicolas


On Fri, Oct 28, 2016 at 6:12 PM, Shayne Fletcher <
shayne.fletcher.50@gmail.com> wrote:

>
> On Fri, Oct 28, 2016 at 9:28 AM, Shayne Fletcher <
> shayne.fletcher.50@gmail.com> wrote:
>
>> On Fri, Oct 28, 2016 at 9:25 AM, Nicolas Ojeda Bar <
>> nicolas.ojeda.bar@lexifi.com> wrote:
>>
>>> You can add a constrain to your functor arguments :
>>>
>>> module type MUL = functor (E : EQ) (N : NUM with type t = E.t) ->
>>>     MUL_S with module N := N and module E := E
>>>
>>
>> ​That does it! I tried so many different things :) You're a rock 'n roll
>> star.
>>
>
> ​Still stuck on how to achieve this I'm afraid.
>
> module type EQ = sig
>   type t
>   val eq : t * t -> bool
> end
>
> module type NUM = sig
>   type t
>   val from_int : int -> t
>   val ( + ) : t -> t -> t
> end
>
> module type MUL_S = sig
>   include EQ
>   include NUM with type t := t
>
>   val mul : t -> t -> t
> end
>
> module type MUL = functor (E : EQ) (N : NUM with type t = E.t) -> MUL_S
>
> module Mul_default (E : EQ) (N : NUM with type t = E.t) : MUL  = struct
>
>   include E
>   include (N : NUM with type t := E.t)
>
>   let mul (x : t) (y : t) : t = failwith "foo"
>
> end
>
> Yields.
>
> ​Error: Signature mismatch:
>        Modules do not match:
>          sig
>            type t = E.t
>            val eq : t * t -> bool
>            val from_int : int -> E.t
>            val ( + ) : E.t -> E.t -> E.t
>            val mul : t -> t -> t
>          end
>        is not included in
>          MUL​
>
> --
> Shayne Fletcher
>

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