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Lennart Augustsson wrote:
Sure, but we also have para f e xs = snd $ foldr (\ x ~(xs, y) -> (x:xs, f x xs y)) ([], e) xs Nice one.
Sure, but we also have
para f e xs = snd $ foldr (\ x ~(xs, y) -> (x:xs, f x xs y)) ([], e) xs Nice one.
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