Veer,
I get this error on ghci : {- `a' is not applied to enough type arguments Expected kind `*', but `a' has kind `* -> *' In the type `SS a' In the type `(Monad a) => {Monad (SS a)}' In the instance declaration for `Monad (SS a)' -}
So, what you are running into is not as much a type error; it's a kind error. Kinds give structure to types, in the same way as types give structure to values. For instance, [Int] and [Maybe Int] are both well-formed types, but [Maybe] is not: Maybe still expects a type argument. Now, let's have a look at kinds. Int is a well-formed type in its own right; we say that it has kind *. (* is pronounced as 'type' or sometimes as 'star'). The type of lists, however, [], is to be applied to a type argument in order to form a well-formed type: so [] has kind * -> *. The same holds for Maybe: it requires a type argument and so it has kind * -> *. Summarizing: Int :: * [] :: * -> * Maybe :: * -> * Now, why is [Maybe] not well-formed? Recall: [] has kind * -> *, so it expects a type argument of kind *. Here, we have supplied as type argument Maybe, which has kind * -> *. (Indeed, [Maybe] is just sugar for [] Maybe.) So, the kind do not match and we are confronted with a kind error. Over to your code snippet.
data SS a = SS a Int
Your type constructor SS expects a single type argument, so we have SS :: * -> * Instances of the Monad type class are to have kind * -> * (for instance, [], Maybe, IO, ...); so, in terms of kinds, SS is a good candidate instance of Monad. But then:
instance (Monad a)=> Monad (SS a) where
Let's see. SS had kind * -> *. This implies that, for SS a to be well- kinded, the type argument a is to be of kind *. But instances of Monad are of kind * -> * and you writing Monad a in the instance head, implies that the type variable a had kind * -> *. Of course, the variable a cannot be of both kind * and kind * -> *. Hence, GHCi nicely presents you a kind error. How to get out of this misery? I'd say, just get rid of the instance head: instance Monad SS where return x = SS x 0 SS x m >>= f = let ~(SS y n) = f x in SS y (m + n) or instance Monad SS where return x = SS x 1 SS x m >>= f = let ~(SS y n) = f x in SS y (m * n) HTH, Stefan