Re: [Caml-list] Default methods for module signatures

[Yotam Barnoy <yotambarnoy@gmail.com>]

[-- Attachment #1: Type: text/plain, Size: 2809 bytes --]

OK so just to clarify, the functor method you suggested does the opposite
of what I was looking for: if my original module (which I wrote to match an
interface) has no 'default function', it adds it, but if my original module
provided that function, it's overridden. What I'm looking for is a solution
whereby if my original module is missing an interface function, that
function is added, but if it already contains a function, the original
function takes precedence.

More specifically regarding my example (just to illustrate the idea): the
Monad module will be used by other functors to do things, so the Monad
interface must define the full interface, including >>.




On Wed, Feb 5, 2014 at 2:43 PM, Martin Jambon <martin.jambon@ens-lyon.org>wrote:

> On Wed 05 Feb 2014 10:49:29 AM PST, Yotam Barnoy wrote:
>
>> Hello List
>>
>> I would like the following feature, and I'm not enough of an expert in
>> module-fu to know if something like this is doable.
>>
>> Suppose I have a module signature of
>>
>> module type Monad = sig
>>   type 'a m
>>   val return : 'a -> 'a m
>>   val (>>=) : 'a m -> ('a -> 'b m) -> 'b m
>>   val (>>) : 'a m -> 'b m -> 'b m
>> end
>>
>> I would like to have a default implementation for (>>), since a simple
>> default implementation is
>>
>> let (>>) m f = m >>= fun _ -> f
>>
>> Alternatively, I would like to include this from some DefaultMonad
>> module, but have the (>>=) referred to in the function be my newly
>> defined (>>=) implementation (ie. late binding). Is there currently
>> any way to do this? If not, would there be a way to implement a
>> partial default implementation built into or associated with a module
>> signature? Something like
>>
>
> OCaml has functors but they don't support optional fields in their
> arguments.
>
> You can create a functor Monad.Make that takes a module without (>>) and
> creates a module with an extra (>>) field. However, if the input module
> contains (>>) already, the new implementation will override it as in this
> minimal example:
>
> # module A = struct end;;
> module A : sig  end
>
> # module M(A: module type of A) = struct include A let x = 0 end;;
> module M : functor (A : sig  end) -> sig val x : int end
>
> # module B = M(struct end);;
> module B : sig val x : int end
>
> # module C = M(struct let x = 1 end);;
> module C : sig val x : int end
>
> # C.x;;
> - : int = 0
>
> C.x is 0, not 1.
>
>
> There may be clever tricks to support some kind of optional module field
> pattern, though. I'll let others scratch their heads. :-)
>
>
>
>  module type Monad = sig... default struct... end
>>
>> Haskell has this available as part of the semantics of their typeclass
>> system, and I think it would be really handy to have (if there isn't
>> already a way to do it currently).
>>
>> -Yotam
>>
>>
>>
>

[-- Attachment #2: Type: text/html, Size: 3805 bytes --]