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

[Brandon S. Allbery KF8NH]

On 2008 Nov 30, at 12:43, Max Rabkin wrote:
> 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.


No disadvantage aside from it making tuples indistinguishable from  
lists.

-- 
brandon s. allbery [solaris,freebsd,perl,pugs,haskell] allbery at kf8nh.com
system administrator [openafs,heimdal,too many hats] allbery at ece.cmu.edu
electrical and computer engineering, carnegie mellon university    KF8NH