Re: [Haskell-beginners] defining 'init' in terms of 'foldr'

On Saturday 04 December 2010 23:20:51, Paul Higham wrote:

Not sure if this thread is still active but I also struggled with this same exercise. I offer the following solution as a thing to shoot at:

myInit :: [a] -> [a] myInit ys = foldr snoc [] $ (\(x:xs) -> xs) $ foldr snoc [] ys where snoc = (\x xs -> xs ++ [x])

init === reverse . tail . reverse only holds for finite lists, for infinite lists xs, reverse xs = _|_, but init xs = xs. Also, it's inefficient, but that's not the point of the exercise.

Note that snoc is defined at the top of the same page as the exercise in Simon's book.

::paul

Cheers, Daniel