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

On Tue, Jan 26, 2010 at 1:56 PM, Ryan Ingram wrote:

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.

This is the only thing I use -XImplicitParams for: {--} :t \x g a -> g (?f a x) \x g a -> g (?f a x) :: (?f::t -> t1 -> t2) => t1 -> (t2 -> t3) -> t -> t3 Which differentiates the givens and gives them names :-) Luke