Re: [Haskell-cafe] instance monad problem

[Creighton Hogg]

Hi

On 5/14/07, Veer Singh [EMAIL PROTECTED] wrote:


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)
___
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe

Re: [Haskell-cafe] instance monad problem

[Stefan Holdermans]

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
___
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe