Will Jones wrote:
> f :: Int -> IO ()
> f = undefined
> g :: Int -> Int -> IO ()
> g = undefined
> h :: Int -> Int -> Int -> IO ()
> h = undefined
vtuple f :: IO (Int -> (Int, ()))
vtuple g :: IO (Int -> Int -> (Int, (Int, ())))
I've tried to type vtuple using a type class; [...]
I've thought about it and it seems impossible to solve this problem
-- you keep needing to ``split'' the function type one arrow further on.
So you need to use recursion to handle the arbitrary deeply nested
arrows in the type of vtuple's argument. I tried it with type families,
but I don't see a reason why functional dependencies should not work.
{-# LANGUAGE FlexibleInstances, TypeFamilies #-}
module VTupleWithTypeFamilies where
We use two type families to handle the two places where the result type
of vtuple changes for different argument types.
type family F a
type family G a r
So the intention is that the type of vtuple is as follows.
class VTuple a where
vtuple :: a -> IO (G a (F a))
The base case:
type instance F (IO ()) = ()
type instance G (IO ()) r = r
instance VTuple (IO ()) where
vtuple = undefined
And the step case:
type instance F (a -> b) = (a, F b)
type instance G (a -> b) r = a -> G b r
instance VTuple b => VTuple (a -> b) where
vtuple = undefined
A test case:
f :: Int -> Bool -> Char -> Double -> IO ()
f = undefined
test = do
vt <- vtuple f
return (vt 5 True 'x' 1.3)
Testing it with ghci yields the following type for test, which looks
good to me.
test :: IO (Int, (Bool, (Char, (Double, ()))))
HTH, Tillmann
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