A simple binary expression tree in Haskell without operator precedence and without parentheses.
Any comments would be much appreciated.
data Tree a= Const a
| X
| Plus (Tree a) (Tree a)
| Mult (Tree a) (Tree a)
| Minus (Tree a) (Tree a)
| Div (Tree a) (Tree a)
| Power (Tree a) (Tree a)
| Cos (Tree a)
| Sin (Tree a)
deriving Show
eval :: ( Eq t, Floating t) => Tree t-> t -> Maybe t
eval (Const a) x = Just (a)
eval X x = Just x
eval (Plus a b) x = (+) <$> eval a x <*> eval b x
eval (Mult a b) x = (*) <$> eval a x <*> eval b x
eval (Minus a b) x = (-) <$> eval a x <*> eval b x
eval (Power a b) x = (**) <$> eval a x <*> eval b x
eval (Div a b) x = if (eval b x) == (Just 0) then Nothing else (/) <$> eval a x <*> eval b x
eval (Cos a) x =(cos) <$> eval a x
eval (Sin a) x = (cos) <$> eval a x
dif :: ( Eq t, Floating t) => Tree t->Tree t
dif (Const a) = Const 0
dif X = Const 1
dif (Plus a b) =Plus (dif a) (dif b )
dif (Mult a b) = Plus (Mult (dif a) b) (Mult a (dif b))
dif (Minus a b) = Minus (dif a) (dif b)
dif (Div a b) = Div (Minus (Mult (dif a) b) (Mult a (dif b))) (Mult b b)
dif (Power (Const a) (Const b)) = Const 0
dif (Power a (Const b)) = Mult (Const b) (Power X (Minus (Const b) (Const 1)))
dif (Cos a) = Minus (Const 0) (Mult (dif a) (Sin a))
dif (Sin a) = (Mult (dif a) (Cos a))
