On Fri, 17 May 2002, Jerry wrote:
Hi, I'm sorry to bother everyone again with this simple append' stuff -- now this is something I _really_ don't understand: -- x is of type [a], [y] is of type [a], and isn't foldr (:) [a] [a] -- perfectly valid?!
:type foldr forall a b. (a -> b -> b) -> b -> [a] -> b now foldr is used with (:) which has type a -> [a] ->[a] so, to find the type of foldr (:) the type (a-> b -> b) must be unified with (:):: c->[c] -> [c] therefore, (replacing a with c and b with [c]) foldr (:) :: [c] -> [c] -> [c] which is what you expected. however, you forgot. 1) append' :: [[a]] -> a -> [[a]] 2) append' [] y = [[y]] 3) append' (x:xs) y = 4) case xs of [] -> foldr (:) [y] x 5) (z:zs) -> (init (x:xs)) ++ [(last xs)++[y]] in "case of [] -> foldr (:) [y] x", foldr (:) [y] x :: [c], the same type as x. but append' works on lists of type [[c]] adds a element of type c and returns something of type [[c]], as you have declared in the type declaration of append'. so, what is the real problem? the type given by foldr (:) [y] x :: [c] doesn't match the resultant type of append', namely [[c]] Also, another good thing to check is that all branches of a case statement have the same type. If something x =case x of True -> 3; False -> "lala" could compile, what would be it's type?? (Hint, if you come from the dynamically typed world and do want to do something of the sort, there is the Either type. case x of True -> Left 3; False -> Right "lala" Jay Cox