On Mon, Sep 22, 2008 at 11:10 PM, Anatoly Yakovenko
type One = S Z type Two = S One etc.
why does:
data Nat a where Z :: Nat a S :: Nat a -> Nat (S a)
data Z data S a
type One = S Z n00 = Z n01::One = S n00
give me:
test.hs:10:11: Couldn't match expected type `One' against inferred type `Nat (S a)' In the expression: S n00 In a pattern binding: n01 :: One = S n00 Failed, modules loaded: none.
or better yet, how is type S Z different from, n01 :: forall a. Nat (S a)
In short, S Z is a type and n01 is a value.
One point that's important to keep in mind is that types and data
constructors have disjoint namespaces, so you can have a type Z and a
data constructor Z which do not need to have any connection.
It may be clearer if we rename the data constructors for Nat.
data Z
data S n
type One = S Z
data Nat :: * -> * where
Zero :: Nat Z
Succ :: Nat n -> Nat (S n)
one :: Nat One
one = Succ Zero
Similarly, One is a type (One :: *) and one is a value (one :: Nat One
and Nat One :: *).
Note also that Z and S are declared here as empty types, using a
common extension. That means there are no (non-bottom) values that
have type Z or S a. This means that Z and S are only used as arguments
to other type constructors or in class contexts.
As long as we're discussing type-level naturals, you may find this old
post of mine interesting.
http://www.haskell.org/pipermail/haskell/2005-May/015815.html
--
Dave Menendez