And this public foldr :: (a -> b -> b) -> b -> [a] -> b public foldr f z [] = z public foldr f z (x:xs) = f x (foldr f z xs)
or is it public foldr :: (a -> b -> b) -> b -> [a] -> b foldr f z [] = z foldr f z (x:xs) = f x (foldr f z xs) and now things aren't lined up. Jared. On 2/24/06, Bulat Ziganshin <[EMAIL PROTECTED]> wrote: > Hello Claus, > > Friday, February 24, 2006, 7:53:09 PM, you wrote: > > CR> public class C a > CR> where > CR> public m1 :: a > CR> private m2 :: a -> String > > please don't stop on this! > > public map (private f) (public (private x:public xs)) = > private (public f (private x)) > `public :` > private map (public f) (private xs) > > > -- > Best regards, > Bulat mailto:[EMAIL PROTECTED] > > _______________________________________________ > Haskell-prime mailing list > Haskell-prime@haskell.org > http://haskell.org/mailman/listinfo/haskell-prime > -- http://www.updike.org/~jared/ reverse ")-:" _______________________________________________ Haskell-prime mailing list Haskell-prime@haskell.org http://haskell.org/mailman/listinfo/haskell-prime
