Hi Andreas,

On Mon, Apr 29, 2002 at 06:48:11PM +0200, Andreas Rossberg wrote:
> 
> Well, if you have a functor like
> 
> 	F : functor(X : sig module type S  module Y:S end) -> ...
> 
> then it would be polymorphic in an unknown number of types.

Perhaps our views differ. What I gathered from Jones' and Russo's
papers was that modules do not contain types. So, the module type
S cannot be a component of X; rather, the type of the functor F
will be universally quantified over S. This leads me to something
like:

  F : forall S. functor (X : S) -> ...

where the distinction between X and Y is eliminated, because it
becomes superfluous. In fact, the `functor' syntax and the name
X are just sugar, since a functor is a function. So I would really
write

  F : forall S. S -> ...

> The application
> 
> 	F (struct module type S = sig type t type u val x : t end 
> 	          module Y = struct type t = int
> 	                            type u = bool
> 	                            val x = 7 end
> 	   end)
> 
> corresponds to something like
> 
> 	F {t:*,u:*} {x:int} {t=int,u=bool} {Y={x=7}}

I would simply apply

  F { x : int } { x = 7 }

or, perhaps (if abstraction is desired)

  F (exists t,u.{ x : t }) (pack { x = 7 } as exists t,u.{ x : t })

So, in this example, we seem to need neither higher kinds nor
kind polymorphism. But perhaps my encoding doesn't have the
features you'd wish?

-- 
François Pottier
Francois.Pottier@inria.fr
http://pauillac.inria.fr/~fpottier/
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