On Wed, Jun 1, 2011 at 3:28 AM, Daniel Fischer <daniel.is.fischer@googlemail.com> wrote:
On Wednesday 01 June 2011 12:25:06, Adrien Haxaire wrote:
>  On Wed, 01 Jun 2011 11:46:36 +0200, Henning Thielemann wrote:
> > Really, you can write foldr in terms of foldl? So far I was glad I
> > could
> > manage the opposite direction.
>foldr (++) (repeat "No way! ")

How about this:

myFoldr :: (a -> b -> b) -> b -> [a] -> b
myFoldr f z xs = foldl' (\s x v -> s (x `f` v)) id xs $ z

Cheers,
Ivan