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
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