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
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