# Foldl as foldr

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## Revision as of 22:01, 16 February 2009

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?

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