Re: [Haskell-cafe] difference between type and newtype

26 Aug 2006


      Andrea Rossato wrote:
...
this is what I'm trying to do, sort of: turn the code at the button
into the do-notation.[1]
module StateOutputMonad where
-- do notation only works with instances of Monad
import Control.Monad
...
data Term = Con Int
         | Add Term Term
           deriving (Show)
type MSO a = State -> (a, State, Output)
-- Use a newtype so you can declare it as a Monad

newtype MSO a = MSO (State -> (a, State, Output))
...
type State = Int
type Output = String
formatLine :: Term -> Int -> Output
formatLine t a = "eval (" ++ show t ++ ") <= " ++ show a ++ " - "
mkMSO :: a -> MSO a
mkMSO a = \s -> (a, s, "")
bindMSO :: MSO a -> (a -> MSO b) -> MSO b
bindMSO m f = \x ->
            let (a, y, s1) = m x in
            let (b, z, s2) = f a y in
            (b, z, s1 ++ s2)
combineMSO :: MSO a -> MSO b -> MSO b
combineMSO m f = m `bindMSO` \_ -> f
--The above 3 functions are replaced by an instance decl
-- combineMSO (ie >>) is the same as the default method

instance Monad MSO where
    return a = MSO (\s -> (a, s, ""))

    (MSO m) >>= f = MSO $ \x ->
                                  let (a, y, s1) = m x in
                                  let MSO y_bz = f a in
                                  let (b, z, s2) = y_bz y in
                                  (b, z, s1 ++ s2)

-- Note the second let is needed to unwrap the newtype
-- Also note you don't need 3 separate let constructs - you could
-- just use one if you like
...
incMSOstate :: MSO ()
incMSOstate = \s -> ((), s + 1, "")
incMSOstate :: MSO ()
incMSOstate = MSO (\s -> ((), s + 1, ""))
...
outMSO :: Output -> MSO ()
outMSO = \x s -> ((),s, x)
-- We need to wrap the function returned by (outMSO x) as
-- a value of (new)type MSO hence:

outMSO :: Output -> MSO ()
outMSO x = MSO (\s -> ((),s, x))

-- You could also have written outMSO = \x -> MSO ... but it's
-- preferable to put the x on the lhs to avoid the dreaded
-- monomorphism restriction
...
evalMSO :: Term -> MSO Int
evalMSO (Con a) = incMSOstate `combineMSO`
                 outMSO (formatLine (Con a) a) `combineMSO`
                 mkMSO a
evalMSO (Add t u) =  evalMSO t `bindMSO` \a ->
                    evalMSO u `bindMSO` \b ->
                    incMSOstate `combineMSO`
                    outMSO (formatLine (Add t u) (a + b)) `combineMSO`
                    mkMSO (a + b)
evalMSO :: Term -> MSO Int
evalMSO (Con a) = do
     incMSOstate
     outMSO (formatLine (Con a) a)
     return a

evalMSO (Add t u) = do
     a <- evalMSO t
     b <- evalMSO u
     incMSOstate
     outMSO (formatLine (Add t u) (a + b))
     return (a + b)
...
-- To be tested with:
-- evalMSO (Add (Con 6) (Add (Con 16) (Add (Con 20) (Con 12)))) 0
-- We need one more thing: a function to run the monad that's wrapped up
-- inside the MSO newtype:

runMSO :: MSO a -> State -> (a, State, Output)
runMSO (MSO f) s = f s

-- Tested with:
-- runMSO (evalMSO (Add (Con 6) (Add (Con 16) (Add (Con 20) (Con 12))))) 0

Regards, Brian.
-- 
Logic empowers us and Love gives us purpose.
Yet still phantoms restless for eras long past,
congealed in the present in unthought forms,
strive mightily unseen to destroy us.

http://www.metamilk.com

Re: [Haskell-cafe] difference between type and newtype

Brian Hulley