[Haskell-cafe] Re: howto tuple fold to do n-ary cross product?

[Max Rabkin]

On Sun, Nov 30, 2008 at 9:30 AM, Luke Palmer <lrpalmer at gmail.com> wrote:
>  cross :: [a] -> [b] -> [(a,b)]
>
> It's just kind of a pain  (you build [(a,(b,(c,d)))] and then flatten
> out the tuples).  The applicative notation is a neat little trick
> which does this work for you.

It seems to me like this would all be easy if (a,b,c,d) was sugar for
(a,(b,(c,d))), and I can't see a disadvantage to that.

--Max