Re: [Haskell-cafe] foldl in terms of foldr

On Tue, 26 Jan 2010, Xingzhi Pan wrote:

Hi,

I am reading Real World Haskell and have some questions about the piece of code implementing foldl in terms of foldr:

-- file: ch04/Fold.hs 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)

The first argument to foldr is of type (a -> b -> a), which takes 2 arguments. But 'step' here is defined as a function taking 3 arguments. What am I missing here? (Partial application? The order of execution?)

http://www.haskell.org/haskellwiki/Foldl_as_foldr

Btw, is there a way I can observe the type signature of 'step' in this code?

http://www.haskell.org/haskellwiki/Determining_the_type_of_an_expression