# Foldl as foldr

### From HaskellWiki

When you wonder whether to choose foldl or foldr you may remember,

that bothfoldl

foldl'

foldr

foldr

foldr

foldl

It holds

foldl :: (a -> b -> a) -> a -> [b] -> a foldl f a bs = foldr (\b g x -> g (f x b)) id bs a

Now the question are:

- How can someone find a convolved expression like this?
- How can we benefit from this rewrite?

## 1 Folding by concatenating updates

Instead of thinking in terms offoldr

g

I find it easier to imagine a fold as a sequence of updates. An update is a function mapping from an old value to an updated new value.

newtype Update a = Update {evalUpdate :: a -> a}

We need a way to assemble several updates.

To this end we define aMonoid

instance Monoid (Update a) where mempty = Update id mappend (Update x) (Update y) = Update (y.x)

Now left-folding is straight-forward.

foldlMonoid :: (a -> b -> a) -> a -> [b] -> a foldlMonoid f a bs = flip evalUpdate a $ mconcat $ map (Update . flip f) bs

foldr

mconcat

mconcat :: Monoid a => [a] -> a mconcat = foldr mappend mempty

mappend

Update

mconcat

foldl

foldl

By the way:

If you use aState

mapAccumL

## 2 foldl which may terminate early

The answer to the second question is:

We can write afoldl

and thus may also terminate on infinite input.

The functionfoldlMaybe

Nothing

Nothing

foldlMaybe :: (a -> b -> Maybe a) -> a -> [b] -> Maybe a foldlMaybe f a bs = foldr (\b g x -> f x b >>= g) Just bs a