On Wed, Jun 1, 2011 at 3:28 AM, Daniel Fischer < daniel.is.fisc...@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
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