instance monad problem
Hello, I am trying to learn haskell , but i am struggling with types , its been around 7 days , it will be very kind if some explain it why this error , i think this is the only stumbling block . I am looking for the comparison on why similar code works , while other code not . 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)' -} Here is the very small code with comments: data SS a = SS a Int data Maybe1 a = Nothing1 | Just1 a instance Monad Maybe1 where (Just1 x) >>= f = f x --^^ this loads fine in ghci -- where as this instance (Monad a)=> Monad (SS a) where (SS x y) >>= f = f (x y) --^^ does not work , so whats the difference , both have type parameters -- something similar works like this : instance (Eq a)=>Eq (SS a) where (SS x y) == (SS b c) = (y == c) && (x == b) Thanks
Hi
On 5/14/07, Veer Singh
Hello, I am trying to learn haskell , but i am struggling with types , its been around 7 days , it will be very kind if some explain it why this error , i think this is the only stumbling block . I am looking for the comparison on why similar code works , while other code not .
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)' -}
Here is the very small code with comments:
data SS a = SS a Int data Maybe1 a = Nothing1 | Just1 a
instance Monad Maybe1 where (Just1 x) >>= f = f x
--^^ this loads fine in ghci
-- where as this instance (Monad a)=> Monad (SS a) where (SS x y) >>= f = f (x y)
--^^ does not work , so whats the difference , both have type parameters
-- something similar works like this : instance (Eq a)=>Eq (SS a) where (SS x y) == (SS b c) = (y == c) && (x == b)
The problem is that you've overspecified the monad SS. Notice that you only had to write instance Monad Maybe1 not instance Monad (Maybe1 a) That's because you declared Maybe1 to only take in one type parameter. SS also takes in only one type parameter, so you're actually telling it that it should make SS a b into a monad, but there is no SS a b. It might help to look at the definition of the State monad in All About Monads http://www.haskell.org/all_about_monads/html/ You'll see that state is defined as newtype State s a =... and they declare instance Monad (State s) not instance Monad (State s a)
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
participants (3)
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Creighton Hogg -
Stefan Holdermans -
Veer Singh