And this works too :

let _ = () |! (fun z -> z);;

while this:

let _ = () |! (fun ?x z -> z);;

doesn't.

It appears that when the mandatory parameter is typed, the inference kernel
doesn't fail, as seen with the unit type. Here's another working example:

let _ = 1 |! (fun ?x (z : int) -> z);;

didier


2012/11/16 Milan Stanojević <milanst@gmail.com>

> >> I don't understand why foo1 is fine and foo2 isn't. I would have
> >> thought that I can use foo2 wherever I can foo1 since it has a
> >> strictly more general type.
> >> Am I missing something obvious here?
> >
> > Its probably due to the optional labels, this works:
> > let y = () |! foo2 ?x:None ?y:None
>
> The question is why foo1 doesn't need this and foo2 does.
>
> --
> Caml-list mailing list.  Subscription management and archives:
> https://sympa.inria.fr/sympa/arc/caml-list
> Beginner's list: http://groups.yahoo.com/group/ocaml_beginners
> Bug reports: http://caml.inria.fr/bin/caml-bugs
>