Sure, but we also have
para f e xs = snd $ foldr (\ x ~(xs, y) -> (x:xs, f x xs y)) ([], e) xs
So I think using para is fine.
-- Lennart
On Feb 12, 2007, at 18:40 , Bernie Pope wrote:
Nicolas Frisby wrote:
Guess this is a tricky choice for a foldr intro, since it requires a
"paramorphism" (see bananas lenses wires etc.)
para :: (a -> [a] -> b -> b) -> b -> [a] -> b
para f e [] = e
para f e (x:xs) = f x xs (para f e xs)
-- note that the original tail of the list (i.e. xs and not xs') is
used in the else-branch
dropWhile' p = para (\x xs xs' -> if p x then xs' else (x:xs)) []
Actually, several people tried to use para, but of course it is not
in the spirit of the challenge :)
Cheers,
Bernie.
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