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

Luke Palmer wrote:

Larry Evans wrote:

...

contains a cross function which calculates the cross product of two lists. That attached does the same but then used cross on 3 lists. Naturally, I thought use of fold could generalize that to n lists; however, I'm getting error:

The type of the function will not involve tuples, since they can be arbitrary length (dynamic-length tuples are not supported in Haskell; we use lists for that).

cross :: [[a]] -> [[a]]

....

This is the sequence function from Control.Monad. cross :: [[a]] -> [[a]] cross = sequence

cross [[1,2],[3,4]] [[1,3],[1,4],[2,3],[2,4]]

Regards, apfelmus