"type" declarations are not first-class; treat them more like macro
expansions. In particular, you cannot make a function polymorphic
over a type declaration.
You can make this typecheck using a "data" or "newtype" declaration for Id:
newtype Id x = Identity x
(or)
data Id x = Identity x
You do need to wrap/unwrap the "Identity" constructor then.
-- ryan
On Fri, Jun 6, 2008 at 3:41 PM, Klaus Ostermann
Why does the code below not pass the type checker?
If I could explictly parameterize y with the type constructor Id (as e.g. in System F), then 'y Id' should have the type Int -> Int and hence "y Id x" should be OK, but with Haskell's implicit type parameters it does not work.
So, how can I make this work?
Klaus ------------
type Id a = a
x :: Id Int x = undefined
y :: (a Int) -> (a Int) y = undefined
test = y x
Error: Couldn't match expected type `a Int' against inferred type `Id Int' In the first argument of `y', namely `x' In the expression: y x In the definition of `test': test = y x