Re: [Caml-list] Q: functors and "has a" inheritance

[Petter Urkedal <paurkedal@gmail.com>]

On 2016-07-06, Jocelyn Sérot wrote:
> Hi Nicolas,
> 
> Thanks fro your answer.
> If i understand correctly, you mean that if i write, say :
> 
> module type S = sig type t val zero: t end
> module type T = sig type t val zero: t end
> module Make (X : S) = (struct type t = X.t * X.t let zero = X.zero, X.zero end : T)
> module M1 = Make (struct type t = int let zero = 0 end)
> module M2 = Make (struct type t = int let zero = 0 end)
> 
> then the compiler will never be able to deduce that M1.t and M2.t are indeed compatible. Am i right ?

Gerd nicely explained how, so I'm just add a note about why:

1. If the module contained a function rather than a plain constant, it
   would be undecidable in general whether the two structures were
   equal.

2. Even if we could (or adopted syntactic equality as an approximation),
   it would break abstraction:  Structures imported from or depending on
   external libraries could be coincidentally equal at some point and
   different after an upgrade.

So, we would be left with a rather ad-hoc rule about how to compare
structures.  The nominal approach taken by OCaml is consistent even if a
bit conservative.  Note that module paths may include functor
applications.