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 <[EMAIL PROTECTED]> 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 > _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
