lol :: [a] -> Nothing | a
lol [] = Nothing
lol [x] = x
lol (x:xs) = if x > lol xs then x else lol xs
listMax [] = Nothing
listMax (x:xs) = Just $ foldr (\x y -> if x > y then x else y) x xs
lol :: [a] -> Nothing | a[a] -> Maybe a. I'm curious as to where you got the idea that this was possible.lol (x:xs) = if x > lol xs then x else lol xsif x > lol xs means that lol xs has to be evaluated before any actual work can be done. Thus it may keep the whole list in memory, since it needs lol xs afterwards as well. The compiler may eliminate the second call to lol xs in which case it's not as bad, but it still wastes stack space because of the second problem, which is that it's not tail recursive.listMax :: [a] -> Maybe a
listMax [] = Nothing
listMax (x:xs) = Just $ iter x xs where
iter m [] = m
iter m (x:xs) = iter (if x > m then x else m) xs
Maybe was because when I looked it up it talked about monads ands Just, neither of which I understand.lol (x:xs) = if x > lol xs then x else lol xsMaybe is introduced at http://learnyouahaskell.com/making-our-own-types-and-typeclasses#type-parameters (but it's probably best to read it more-or-less in order, at least for the first few chapters).Ord a =>
a has to be an instance of Ord, that is, it must be comparable.
sum. It would be more or less equivalent to using foldl' though. Also, there's nothing wrong with foldl' and seq (and $! and bang patterns).ghc -O runs >>2 and >>3 quite happily. Excessive laziness has rarely caused performance problems in my code, and it's usually pretty easy to fix when it does.
free/delete?