Hi David, ah yes, you're right of course. Interestingly, the Reason
formatting tool reformats both of these into:

let g1 = (x) => {
  print_int(x);
  print_int
};

let g2 = (x) => {
  print_int(x);
  (y) => print_int(y)
};

I think it makes it very easy to tell that the latter just includes a
redundant eta abstraction and can be simplified to the former.

Regards,

Yawar


On Wed, Dec 13, 2017 at 12:20 PM, David Allsopp <dra-news@metastack.com>
wrote:

> Yawar Amin wrote:
> > On Wed, Dec 13, 2017 at 11:38 AM, Marshall <mailto:marshall@logical.net>
> wrote:
> > > On Dec 13, 2017, at 4:37 AM, David Allsopp <mailto:
> dra-news@metastack.com> wrote:
> > >
> > > > In terms of why you might want to be "hiding" currying, or at least
> making it need
> > > > less immediate explanation when learning the language, consider, for
> example,
> > > > explaining the difference between:
> > > >
> > > > let f x y = print_int x; print_int y in
> > > > f 42
> > > >
> > > > and
> > > >
> > > > let g x = print_int x; print_int in
> > > > g 42
>
> (code amended to give the functions different names)
>
> > > Apart from ReasonML, this is interesting.  Why do these behave
> differently?
> > > I don’t understand yet.
> >
> > In the second example, the second `print_int` never gets called.
>
> No, both functions have the same type (int -> int -> unit) and if they're
> applied to two integers then each will print both of them to stdout. The
> difference is when they're partially applied - [f] prints nothing; [g]
> prints 42. It's slightly less unclear in lambda form:
>
> let f =
>   fun x ->
>     fun y ->
>       print_int x;
>       print_int y
>
> let g =
>   fun x ->
>     print_int x;
>     fun y ->
>       print_int y
>
>
> David
>
>