Hello, on my journey through Haskell I've hit a problem where I don't know any further and would greatly appreciate your help. I've got this code so far: data Value a e = VConst a | VFunc (e -> a) evalV :: Value a e -> e -> a evalV (VConst x) = const x evalV (VFunc f) = f liftV :: (a -> a -> t) -> (Value a e) -> (Value a e) -> e -> t liftV f x y e = (evalV x e) `f` (evalV y e) (.==.), (./=.) :: (Eq a) => (Value a e) -> (Value a e) -> e -> Bool (.==.) = liftV (==) (./=.) = liftV (/=) (.&&.), (.||.) .... some more operators This allows me to write client code like this: data Person = Person { personName :: String , personAge :: Int } deriving (Show) exampleExpr :: Bool exampleExpr = (VConst 99) .==. (VFunc personAge) $ Person "pete" 99 I was wondering, whether it'd be possible to enable defining expression without the Value data constructors, i.e. 99 .==. personAge $ Person "pete" 99 I've tried using a type class to have a function for lifting different types into Values. class ToValue a e | a -> e where valueLift :: a -> Value a e then I tried defining an instance for functions: instance ToValue ((->) Person a) Person where valueLift f = undefined in ghci:
:t valueLift personName valueLift personName :: Value (Person -> String) Person
and suddenly I realized this will not work because I would need the type `Value String Person'. Now, I'm stuck and don't know which path to take in order to come closer to being able writing "99 .==. personAge" or "personAge .==. 99". Maybe I'm approaching it in a completely wrong way or it's simply not possible. Do you have any ideas, hints? Many thanks in advance for any kind of feedback, pete.