Prelude> :t foldl (\x -> \xs -> xs:x) [] foldl (\x -> \xs -> xs:x) [] :: [b] -> [b] Strange choice of names, though, since x is a list, and xs is an element. I would have gone for: foldl (\xs x -> x:xs) [] although the library opts for: foldl (flip (:)) [] On 21/09/2007, Miguel Mitrofanov <miguelimo38@yandex.ru> wrote:

...

reverse = foldl (\x -> \xs -> xs:x) []

Doesn't typecheck. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe