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

On 11/23/08 13:52, Luke Palmer wrote:

2008/11/23 Larry Evans :

...

http://www.muitovar.com/monad/moncow.xhtml#list

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:

You should try writing this yourself, it would be a good exercise. To begin with, you can mimic the structure of cross in that tutorial, but make it recursive. After you have a recursive version, you might try switching to fold or foldM.

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]]

However, list's contain elements all of the same type. What the following boost post: http://thread.gmane.org/gmane.comp.lib.boost.devel/182797/focus=182915 demonstrated was, AFAICT, the c++ template metaprogramming counterpart to the moncow haskell cross. Now, AFAICT, the boost vault directory: http://www.boostpro.com/vault/index.php?PHPSESSID=ab51206c9d980155d142f5bcef8e00ee&direction=0&order=&directory=Template%20Metaprogramming in the cross_nproduct_view_test.zip, contains what I'm looking for in haskell. I'm guessing that: templatestruct row_view; corresponds to the haskell tuple type (row,column) I'm trying to confirm that by printing out the typename in a formated form, but I'm having trouble doing that at the moment: http://preview.tinyurl.com/66x4nx Is there some version of haskell, maybe template haskell, that can do that, i.e. instead of: cross::[[a]] -> [[a]] have: crossn::[a0]->[a1]->...->[an] -> [(a0,a1,...,an)] ?