Re: [Caml-list] Records typing

[Jacques Garrigue <garrigue@kurims.kyoto-u.ac.jp>]

From: Andreas Rossberg <rossberg@ps.uni-sb.de>
> malc wrote:
> > 
> > # module type T = sig type t type r = { f : t } end;;
> > module type T = sig type t and r = { f : t; }  end
> > 
> > # module M : T with type t = float = struct
> >   type t = float
> >   type r = { f : t }
> > end;;
> > 
> > Signature mismatch: ...
> > 
> > Is there a way to work around this?
> 
> This seems to be a bug. You can work around it using "and" to connect
> the type declarations in the structure:
> 
> # module type T = sig type t type r = { f : t } end;;
> module type T = sig type t and r = { f : t; }  end
> # module M : T with type t = float = struct
>   type t = float
>   and r = { f : t }
> end;;
> module M : sig type t = float and r = { f : t; }  end

Actually this is not a bug. And your workaround is very interesting
(and useful).
The fact is, the two definitions are actually semantically different
(at least from the point of view of the compiler).

% ocaml -dlambda
        Objective Caml version 3.06+16 (2002-11-04)

# module M = struct type t = float and r = {f:t} end;;
module M : sig type t = float and r = { f : t; }  end
# {M.f = 1.0};;
[0: 1.0]
- : M.r = {M.f = 1.}
# module M' = struct type t = float type r = {f : t} end;;
module M' : sig type t = float and r = { f : t; }  end
# {M'.f = 1.0};;
[|1.0|]
- : M'.r = {M'.f = 1.}

In the first case, the record is represented as an array of a pointer
to a boxed float, while in the second case it is represented as an
array of an unboxed float. Since the data representation is different,
these two signatures are not equivalent. I was afraid there would be
no direct way to express the first case, but actually you can.

Jacques Garrigue
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