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

On Sun, Nov 30, 2008 at 9:30 AM, Luke Palmer 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