Bernie Pope wrote: > 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.
"Nice one" is an euphemism, it's exactly solution one :) Regards, apfelmus _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
