Xingzhi Pan wrote:

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?

You can think of step as a function of two arguments which returns a function with one argument (although in reality, as any curried function, 'step' is _one_ argument function anyway): step :: b -> (a -> c) -> (b -> c) e.g. 'step' could have been defined as such: step x g = \a -> g (f a x) to save on lambda 'a' was moved to argument list. -- View this message in context: http://old.nabble.com/foldl-in-terms-of-foldr-tp27322307p27324376.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.

On Tue, Jan 26, 2010 at 5:04 AM, Xingzhi Pan wrote:

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)

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

Here is how I do it: Prelude> :t \[f] -> \x g a -> g (f a x) \[f] -> \x g a -> g (f a x) :: [t1 -> t -> t2] -> t -> (t2 -> t3) -> t1 -> t3 The [] are unnecessary but help me differentiate the "givens" from the actual arguments to the function. Here's a way that gives better type variables and properly constrains the type of f: Prelude> let test [f] x g a = g (f a x) where typeF = f `asTypeOf` const Prelude> :t test test :: [a -> b -> a] -> b -> (a -> t) -> a -> t -- ryan