Repeated thanks to you, Adam! Your code is brilliantly simple. Sadly, I cannot reproduce the behaviors in your comments on my ghci (7.6.1) ..... Can we guess why? The version of packages we are using? Mines are here. https://github.com/nushio3/practice/tree/master/variable-arity/adam

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:t forZ [1,2,3] (+) forZ [1,2,3] (+) :: (Num t, Num a, TypeCast br HFalse, HBuild2' br (HCons [t] HNil) (a -> a -> a) r) => r forZ [1,2,3] [10] (+)

<interactive>:13:1: Couldn't match type `[y]' with `(a0 -> a0 -> a0) -> t0' When using functional dependencies to combine Apply ApplyZap (a, b) [y], arising from the dependency `f a -> r' in the instance declaration at Part1.lhs:193:12 Apply ApplyZap ([[t2]], [t1]) ((a0 -> a0 -> a0) -> t0), arising from a use of `forZ' at <interactive>:13:1-4 In the expression: forZ [1, 2, 3] [10] (+) In an equation for `it': it = forZ [1, 2, 3] [10] (+)

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forZ [1,2,3] "hi there" (,)

<interactive>:14:1: Couldn't match type `[y]' with `(a0 -> b0 -> (a0, b0)) -> t0' When using functional dependencies to combine Apply ApplyZap (a, b) [y], arising from the dependency `f a -> r' in the instance declaration at Part1.lhs:193:12 Apply ApplyZap ([[Char]], [t1]) ((a0 -> b0 -> (a0, b0)) -> t0), arising from a use of `forZ' at <interactive>:14:1-4 In the expression: forZ [1, 2, 3] "hi there" (,) In an equation for `it': it = forZ [1, 2, 3] "hi there" (,) Best, Takayuki 2012/12/11 adam vogt :

On Sat, Dec 8, 2012 at 10:27 AM, Takayuki Muranushi wrote:

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Continued discussion from

https://groups.google.com/d/topic/haskell-cafe/-e-xaCEbd-w/discussion https://groups.google.com/d/topic/haskell-cafe/kM_-NvXAcx8/discussion

Thank you for all the answeres and thinkings;

Here's zipWithN for general Zip functors: [1] . This, together with [2] may constitute a small hackage. A modification from Wren's idea to [1] is the use of fmap instead of repeat.

I'm wondering if there are any laws for Zip functors. I first thought that there are similarity between Zips and Applicatives, as [3] states

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instance Applicative f => Zip f where zip = liftA2 (,)

However, my intuition is that zipping two arrays should result in an array of size of the same order as two, giving rise to a Zip functor law candidate:

zipWith const xs $ zipWith const xs ys == zipWith const xs ys

which is violated by the above statement "zip = liftA2 (,)" .

[1] https://github.com/nushio3/practice/blob/master/variable-arity/ZipWithN-2.hs [2] https://github.com/nushio3/practice/blob/master/free-objects/zipf-12.hs [3] http://hackage.haskell.org/packages/archive/TypeCompose/0.9.7/doc/html/Data-...

Hi again, Takayuki

While the forZN in zipf-12 is able to infer the result type given arguments, it doesn't give any useful information about types for arguments unlike an example here:

http://code.haskell.org/~aavogt/flip_zipWithN/P4.hs

which imports a slight modification of Paczesiowa's code: http://code.haskell.org/~aavogt/flip_zipWithN/Part1.lhs

But maybe it isn't possible to infer much about earlier arguments given later ones since there is an instance Zip ((->) a), that forZN apparently can work with.

Adam

-- Takayuki MURANUSHI The Hakubi Center for Advanced Research, Kyoto University http://www.hakubi.kyoto-u.ac.jp/02_mem/h22/muranushi.html