Re: [Haskell-beginners] Is there an "unscan" function?

Am 13.01.2012 19:19, schrieb Stephen Tetley:

Hi Jeffrey

My version actually contains an error, it should be:

unscan :: (a -> a -> b) -> [a] -> [b] unscan f (a:b:bs) = f a b : unscan f (b:bs) unscan _ _ = []

A slightly less concise, but optimized version avoids putting the second element back in a list:

unscan f (a:b:bs) = f a b : go b bs where go _ [] = [] go x (z:zs) = f x z : go z zs unscan _ _ = []

This version contains duplicate code: The first line could be: unscan f (a : bs) = go a bs "putting the second element back" can be avoided by @-Patterns! unscan f (a : bs@(b : _)) = f a b : unscan bs Cheers Christian