Re: [Haskell-beginners] Structural restrictions in type constructor

23 Jun 2015

      Imants,

You are right. The problem is not in IO here, it's that if you have access
to data-constructor, you can do things like:

six :: TypeValInt 6
six = TypeValInt 5

Initially, I was making an assumption that you won't be using a
data-constructor. After thinking about it a bit, I should note that my code
isn't much different from just using a "smart constructor" approach, e.g.
hiding a real MyP constructor, and instead providing a function:

mkMyP (a, b) = MyP (a, b) (b, a)

and exporting only this function. This would make sure all your users only
create a valid set of data.

On Tue, Jun 23, 2015 at 5:05 PM, Imants Cekusins  wrote:
...
On 23 June 2015 at 14:54, Kostiantyn Rybnikov  wrote:
...
Hi Matt. I don't know how bad is this, but here's what I came up with.
...
this modified for IO version accepts any input, including that which
should have caused error:
or did I do something wrong?
{-# LANGUAGE DataKinds #-}
{-# LANGUAGE KindSignatures #-}
{-# LANGUAGE ExistentialQuantification #-}
{-# LANGUAGE PolyKinds #-}
{-# LANGUAGE StandaloneDeriving #-}
module PairsMatchedKR where
import GHC.TypeLits
data TypeValInt (n::Nat) = TypeValInt Int
    deriving (Show)
one :: TypeValInt 1
one = TypeValInt 1
two :: TypeValInt 2
two = TypeValInt 2
data MyP a b = MyP (TypeValInt a, TypeValInt b) (TypeValInt b, TypeValInt
a)
    deriving (Show)
main :: IO ()
main = do
    putStrLn "Hello!"
    x1 <- getLine
    x2 <- getLine
    x3 <- getLine
    x4 <- getLine
print (MyP (tvi x1, tvi x2) (tvi x3, tvi x4))
class TypeVal (g :: a -> *)
instance TypeVal TypeValInt
data MyPGen a b = forall g. (TypeVal g, Show (g a), Show (g b))
               => MyPGen (g a, g b) (g b, g a)
deriving instance Show (MyPGen a b)
tvi:: String -> TypeValInt (n::Nat)
tvi = TypeValInt . read
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