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Lennart Augustsson wrote:
.........para f e xs = snd $ foldr (\ x ~(xs, y) -> (x:xs, f x xs y)) ([], e) xs I thought solution one was missing the ~ ?
......para f e xs = snd $ foldr (\ x ~(xs, y) -> (x:xs, f x xs y)) ([], e) xs
...para f e xs = snd $ foldr (\ x ~(xs, y) -> (x:xs, f x xs y)) ([], e) xs
para f e xs = snd $ foldr (\ x ~(xs, y) -> (x:xs, f x xs y)) ([], e) xs
I thought solution one was missing the ~ ?
Yes, that's irrefutably right ;) I mean solution one modulo the laziness bug. Regards, apfelmus
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