"ocaml_beginners"::[] Typing problem: qualifying constructor

Hello,

I have changed the stdlib Map file and added the following type:

type 'a enum =
| End
| More of 'a * 'a t * 'a enum

and have made it "visible" (i.e: non-abstract) in both the interface and
the functor. In the "client" code I would like to use "End". To do this
I simple use:

module Key_map = Imap.Make (struct
type t = dt_key
let compare = Pervasives.compare
end)

and "access" End so:

let iter = Key_map.End

My question is: assume I had another visible type within the functor
that also had a "End" constructor. How could I "qualify" Key_map in
order specifically use the type 'a enum?

TIA,
Hugo F.

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.

On Tue, May 22, 2007 at 10:35:36AM +0100, Hugo Ferreira wrote:
> My question is: assume I had another visible type within the functor
> that also had a "End" constructor. How could I "qualify" Key_map in
> order specifically use the type 'a enum?

If the two constructors end up in the same module then you can't. The
second one "wins".

You could assign End to a value and use that instead, eg, in the
functor, add this before the second definition:

let first_end = End

but really you're better off not having two constructors with the same
name.

Rich.

--
Richard Jones
Red Hat

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.

Richard,

Richard Jones wrote:
> On Tue, May 22, 2007 at 10:35:36AM +0100, Hugo Ferreira wrote:
>> My question is: assume I had another visible type within the functor
>> that also had a "End" constructor. How could I "qualify" Key_map in
>> order specifically use the type 'a enum?
>
> If the two constructors end up in the same module then you can't. The
> second one "wins".
>

I see.

> You could assign End to a value and use that instead, eg, in the
> functor, add this before the second definition:
>
> let first_end = End
>

Haaaa.. good one. Won't work for constructor with values though. 8-(

> but really you're better off not having two constructors with the same
>; name.
>

Hmmmm... sound advice.

Thanks,
Hugo F.

> Rich.
>

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.

On Tue, May 22, 2007 at 05:05:55PM +0100, Hugo Ferreira wrote:
> Richard Jones wrote:
> > let first_end = End
> Haaaa.. good one. Won't work for constructor with values though. 8-(

It can do:

let first_end x y = End (x, y)

Rich.

--
Richard Jones
Red Hat

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.

Richard Jones wrote:
> On Tue, May 22, 2007 at 05:05:55PM +0100, Hugo Ferreira wrote:
>> Richard Jones wrote:
>>> let first_end = End
>> Haaaa.. good one. Won't work for constructor with values though. 8-(
>
> It can do:
>
> let first_end x y = End (x, y)
>

Yes you can. Thanks a lot.
I will use different constructors though.

> Rich.
>

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.