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 bug. Regards, apfelmus _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
