[Haskell-cafe] Re: Foldr tutorial, Inspired by Getting a Fix from a Fold

apfelmus at quantentunnel.de apfelmus at quantentunnel.de
Tue Feb 13 04:14:20 EST 2007

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 ~ ?

Yes, that's irrefutably right ;) I mean solution one modulo the laziness


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