On Aug 19, 2012, at 5:29 , wren ng thornton wrote:

On 8/17/12 5:35 AM, TP wrote:

...

Hi,

I am currently reading documentation on Generalized Algebraic Data Types:

http://en.wikibooks.org/wiki/Haskell/GADT

I have a question concerning this page. Let us consider the following code proposed in the page:

---------------------------------- -- Phantom type variable a (does not appear in any Expr: it is just a -- dummy variable). data Expr a = I Int | B Bool | Add (Expr a) (Expr a) | Mul (Expr a) (Expr a) | Eq (Expr a) (Expr a) deriving (Show) [...] I don't understand. When we write "eval (I n) = n", as I is a constructor which takes an Int as argument, we are able to deduce that the type of n is Int; so the type of eval should be in this case "Expr Int -> Int". What do I miss?

Perhaps it'd help to rewrite the above ADT using GADT syntax (but note that its the exact same data type):

data Expr :: * -> * where I :: Int -> Expr a B :: Bool -> Expr a Add :: Expr a -> Expr a -> Expr a Mul :: Expr a -> Expr a -> Expr a Eq :: Expr a -> Expr a -> Expr a

But if you use the real power of GADT's you would write: {-# LANGUAGE KindSignatures, GADTs #-} data Expr :: * -> * where I :: Int -> Expr Int B :: Bool -> Expr Bool Val :: a -> Expr a Add :: Expr Int -> Expr Int -> Expr Int Mul :: Expr Int -> Expr Int -> Expr Int Eq :: Eq a => Expr a -> Expr a -> Expr Bool eval :: Expr a -> a eval (I i) = i eval (B b) = b eval (Val v) = v eval (Add e1 e2) = eval e1 + eval e2 eval (Mul e1 e2) = eval e1 * eval e2 eval (Eq e1 e2) = eval e1 == eval e2

Note that the I and B cases are actually superfluous, and are covered by the val case. This has all nothing to do with phnatom types, but with a proper understanding of the GADT concept. Doaitse

So, when looking at the pattern (I n), since I :: Int -> Expr a we know that n :: Int and that (I n) :: Expr a.

-- Live well, ~wren

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