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

Tuples would still be distinguishable from lists, since "cons" changes their type: (b,c,d) and (a,b,c,d) would have different types, while [b,c,d] and [a,b,c,d] wouldn't. On 30 Nov 2008, at 20:48, Brandon S. Allbery KF8NH wrote:

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

...

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.

No disadvantage aside from it making tuples indistinguishable from lists.

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

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