I'm no expert, but it looks like the generalization of that would be
some f that took a list:
f :: [a] -> b
so what you'd have is a fold, right?
foldr1 :: (a -> a -> a) -> [a] -> a
Best,
Philip Neustrom
On Sun, Aug 10, 2008 at 11:47 AM, Michael Feathers
I wrote this function the other day, and I was wondering if I'm missing something.. whether there is already a function or idiom around to do this.
unlist3 :: (a -> a -> a -> b) -> [a] -> b unlist3 f (x:y:z:xs) = f x y z
I was also wondering whether the function can be generalized to N or whether this is just one of those edges in the type system that you can't abstract over.
Thanks,
Michael
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