Great!

That was indeed really helpfull! I really missed out that last bit, I
sould have payed more attention.

'a Mon.m and 'a O.m are monadic values. O.m and Mon.m are the same
type.  My intuition would be that they should be variant  ((a :> b m)
should behave like a >>= (fun x -> return (x:>b) where a is a value m
a monadic type and b a supertype of a)) can anyone concur that.

Cheers,
Till

On 3/12/07, Chris King <colanderman@gmail.com> wrote:
> On 3/12/07, Till Varoquaux <till.varoquaux@gmail.com> wrote:
> > This expression cannot be coerced to type
> >  O.o Ast.T.expr O.m = O.o Ast.T.expr Mon.m;
> > it has type O.o Ast_js.T.expr Mon.m but is here used with type
> >  O.o Ast.T.expr O.m = O.o Ast.T.expr Mon.m
> > Type
> >  O.o Ast_js.T.expr =
> >    [ `Binop of To.binop * To.expr * To.expr
> >    | `Cst of To.constant
> >    | `Ecall of To.ident * To.arg list
> >    | `Fun of To.ident list * To.bloc
> >    | `Lval of To.lvalue
> >    | `Unop of To.unop * To.expr ]
> > is not compatible with type
> >  O.o Ast.T.expr =
> >    [ `Binop of To.binop * To.expr * To.expr
> >    | `Cst of To.constant
> >    | `Ecall of To.ident * To.arg list
> >    | `Epos of Ast.location * To.expr
> >    | `Fun of To.ident list * To.bloc
> >    | `Lval of To.lvalue
> >    | `Unop of To.unop * To.expr ]
> > The first variant type does not allow tag(s) `Epos
>
> What sort of type is 'a O.m (or 'a Mon.m)?  If it's something whose 'a
> member is mutable, then such a coercion will be impossible (it would
> allow you to set the member to have the `Epos tag, which it can't).
> This would also happen if 'a O.m is declared as an abstract type,
> since O'Caml assumes invariance by default.  In the latter case, you
> need to specify that 'a O.m is covariant with its parameter:
>
> type +'a O.m
>
> This tells Caml that if 'a is a subtype of 'b, then 'a O.m is indeed a
> subtype of 'b O.m and you will be allowed to perform the above
> coercion.
>
> Hope this helps.
>
> - Chris
>