Pankaj Godbole's code examples for 8.9.6 are sufficiently more complex than mine,that I am worried I missed an obvious bug I'm not experience enough to know about or test for. Correction or reassurances would be appreciated. Regards, Trent {-- 5. Given the type declaration data Expr = Val Int | Add Expr Expr define a higher-order function folde :: (Int -> a) -> (a -> a -> a) -> Expr -> a such that folde f g replaces each Val constructor in an expression by the function f, and each Add constructor by the function g. Hutton, Graham. Programming in Haskell (Kindle Locations 3364-3371). Cambridge University Press. Kindle Edition. --} data Expr = Val Int | Add Expr Expr -- Trent's stuff folde :: (Int -> a) -> (a -> a -> a) -> Expr -> a folde f g (Val x) = f x folde f g (Add left right) = g (folde f g left) (folde f g right) {-- 6. Using folde, define a function eval :: Expr -> Int that evaluates an expression to an integer value, and a function size :: Expr -> Int that calculates the number of values in an expression. Hutton, Graham. Programming in Haskell (Kindle Locations 3372-3374). Cambridge University Press. Kindle Edition. --} -- Trent's stuff eval :: Expr -> Int eval = folde id (+) size :: Expr -> Int size = folde (\_ -> 1) (+) -- hlint suggests \_ as const 1 test0 :: Expr test0 = Val 1 test1 :: Expr test1 = Add (Val 1) (Val 2) test3 :: Expr test3 = Add (Val 1) (Add (Val 2) (Val 3)) -- the below are from: -- Pankaj Godbole -- https://github.com/pankajgodbole/hutton/blob/master/exercises.hs evalE :: Expr -> Int evalE (Val n) = n evalE (Add e1 e2) = folde toEnum (+) (Add e1 e2) -- doesn't a Val have to prefix an int? Then why is f "toEnum"? numVals :: Expr -> Int numVals (Val n) = 1 numVals (Add e1 e2) = numVals e1 + numVals e2