[Haskell-cafe] Re: howto tuple fold to do n-ary cross product?
max.rabkin at gmail.com
Sun Nov 30 12:43:19 EST 2008
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.
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