Question

Consider the following code,

data MyBaseExpr α where
    ConstE :: Show α => α -> MyBaseExpr α

class Monad  => MyMonadCls  where
    type ExprTyp  :: * -> *
    var :: String -> ExprTyp  α ->  (ExprTyp  α)

expTypArg :: forall  α. MyMonadCls  => ExprTyp  α ->  α
expTypArg a = undefined

-- dummy type which will be used as an instance
newtype A  α = A ( α)

Then, if one tries to write an instance using the expTypeArg function,

instance forall . (Monad , Monad (A )) => MyMonadCls (A ) where
    type ExprTyp (A ) = MyBaseExpr
    var nme init@(expTypArg -> typb) =
        return init

the compiler complains

Couldn't match type `ExprTyp 0' with `MyBaseExpr'
Expected type: ExprTyp (A ) α
  Actual type: ExprTyp 0 α

But, if one adds some scoped type expressions,

instance forall . (Monad , Monad (A )) => MyMonadCls (A ) where
    type ExprTyp (A ) = MyBaseExpr
    var nme init@((expTypArg :: MyMonadCls (A ) =>
                   ExprTyp (A ) α ->
                   (A  α)) -> typb) =
        return init

then it resolves fine. What’s the problem resolving ExprTyp == MyBaseExpr for expTypArg?

edit

Thanks very much, Daniel! Here’s a way to take some of the verbage out, after noticing that only the type of need be enforced.

ignore_comp ::  α ->  β ->  β
ignore_comp a b = b

instance forall . (Monad , Monad (A )) => MyMonadCls (A ) where
    type ExprTyp (A ) = MyBaseExpr
    var nme init@(expTypArg -> typb) =
        typb `ignore_comp`
        return init

Answer 1

ExprTyp is not (necessarily) an injective type function. This means that being handed something of type ExprType m doesn’t nail down m — there may also be a different n such that ExprType n = ExprType m. This makes the type of expTypArg a bit tricky: it uses return-type polymorphism in the same way, e.g., read does, so you’ll need to give extra type annotations to its result in the same situations you have to do with read.