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 lin
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 :`
p
Hello Claus,
Friday, February 24, 2006, 6:55:51 PM, you wrote:
CR> not quite (though I believe that would be close to Simon M's idea).
CR> in my modification, both map and length would move completely
CR> into the export section
WHY? it's not the interface. implementation of exported functions
