Federico Mastellone wrote:
foldl :: (a -> b -> a) -> a -> t b -> a
Them, outside of the class, foldr' is defined like this:
-- | Fold over the elements of a structure, -- associating to the right, but strictly. foldr' :: Foldable t => (a -> b -> b) -> b -> t a -> b foldr' f z0 xs = foldl f' id xs z0 where f' k x z = k $! f x z
I don't understand this definition, foldl receives 3 parameters and here it is used with 4, how is it possible? Even the function passed to foldl has 3 parameters when a function of 2 is needed.
In the type of foldl, the parameter 'a' can be any type - even a function. So let's see what we get when we substitute 'c -> d' for 'a' in the type of foldl: ((c -> d) -> b -> (c -> d)) -> (c -> d) -> t b -> (c -> d) Now, remembering that -> is right-associative in type expressions, we can remove some parentheses: ((c -> d) -> b -> c -> d) -> (c -> d) -> t b -> c -> d So when 'a' is a function, we see that foldl indeed takes 4 parameters, and its first parameter is a function that takes 3 parameters. Regards, Yitz