Defining foldl in terms of foldr

Question

myFoldl :: (a -> b -> a) -> a -> [b] -> a

myFoldl f z xs = foldr step id xs z
    where step x g a = g (f a x)

I am currently readi

Answer 1

As Adam Smith says in the comments, there are only three arguments to foldr. The line in question gets parsed like this:

myFoldl f z xs = (foldr step id xs) z

There are other implicit brackets too of course, but these are the important ones.

Here is a rewrite with the type annotations, assuming scoped type variables (i.e. a and b mean the same types throughout this definition).

myFoldl :: (a -> b -> a) -> a -> [b] -> a
myFoldl f z xs =  goFold z
   where
      goFold :: a -> a
      goFold = foldr step id xs
      step :: b -> (a -> a) -> (a -> a)  -- Last brackets are redundant
      step x g a = g (f a x)

I've moved the foldr invocation into a separate value goFold so you can see its type and how it gets applied to the z value. The step function accumulates the b values into a function of type (a -> a). Each b value processed by goFold adds an extra one of these. The "zero" for functions is of course id from the Prelude:

id :: a -> a
id x = x

The -> function operator in the types is right associative, so the last pair of brackets in the step type are redundant. But I've written it like that because it higlights the way in which step is being used; it takes a value and a function and returns a new function.