Re: [Haskell-cafe] Type union

Hello, I have a somewhat similar problem, trying to achieve union types but for the purpose of defining the set of allowable outcomes of a function. I have tried naively to define a type operator a :| b and I would like to be able to define a function like : f :: Int -> Int :| String :| Bool f 1 = in 1 1 f 2 = in 2 "foo" f 3 = in 3 True e.g. the codomain type is indexed by integers. I think this should be possible, even without full dependent typing given that I only expect to use literals My type-level-fu is really lacking so help would be appreciated. Regards, -- Arnaud Bailly FoldLabs Associate: http://foldlabs.com On Mon, Sep 15, 2014 at 2:56 PM, Richard Eisenberg wrote:

Have you tried using closed type families with the type-level (==) operator from GHC 7.8's Data.Type.Equality? That's how I've done unions before. The key step is to use a closed type family to write a type family equation that triggers when two types do *not* equal.

Let me know if you need more info...

Richard

On Sep 14, 2014, at 10:32 AM, Dmitry Bogatov wrote:

...

Hello!

I am trying to create type safe boolean formula representation. Main operation is substion of particular value, to get another formula, and function that accept formula to calculate it's value.

So target is:

a = Conjunction (Var X) (Var Y) is 2 variable formula value (apply (X, True) . apply (Y, True) $ a) -- True

and neither

value a apply (X, True) . apply (X, False) $ a

typechecks.

How can I archive it?

class Union a b c instance (a ~ b) => Union a b b instance Union a b (a, b)

data P = P deriving Show data Q = Q deriving Show class (Show a) => Variable a instance Variable P instance Variable Q

data Formula t where Prop :: (Variable b) => b -> Formula b Conjunction :: (Union t1 t2 t3) => Formula t1 -> Formula t2 -> Formula t3

deriving instance Show (Formula t) main = print (Conjunction (Prop P) (Prop Q) :: Int)

This complains on ambitious t3.

Attempt by typefamilies fails, since

type family Union t1 t2 :: * data Void type instance Union a a = a type instance Union (a, b) a = (a, b) type instance Union (a, b) b = (a, b) type instance Union (a, b) c = (a, b, c) type instance Union a (a, b) = (a, b) type instance Union b (a, b) = (a, b) type instance Union c (a, b) = (a, b, c) type instance Union (a, b, c) a = (a, b, c) type instance Union (a, b, c) b = (a, b, c) type instance Union (a, b, c) c = (a, b, c) type instance Union (a, b, c) d = Void type instance Union a (a, b, c) = (a, b, c) type instance Union b (a, b, c) = (a, b, c) type instance Union c (a, b, c) = (a, b, c) type instance Union d (a, b, c) = Void

this is somewhy conflicting instances. I am out of ideas.

-- Best regards, Dmitry Bogatov , Free Software supporter, esperantisto and netiquette guardian. GPG: 54B7F00D _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

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