[Haskell-cafe] reverse with foldl

PR Stanley prstanley at ntlworld.com
Fri Sep 21 01:55:13 EDT 2007

reverse = foldl (xs x . x:xs) [] (xs ++ ) = foldl (\ys y -> ys ++ [y]) xs
If I were to define reverse with foldl I would do it this way:
reverse = foldl (\x -> \xs -> xs:x) []

Any idea what the first code frag is suposed to achieve?
Thanks, Paul

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