> > 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 >
Great! Now I really can say "Come on! It's fun! I can write foldr with foldl!" _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
