Hi Johnaton,
Ah yes. That makes sense. Is there a way to define type r to be all types
except functions?
-John
On Fri, Feb 13, 2009 at 10:44 AM, Jonathan Cast
On Fri, 2009-02-13 at 10:34 +1100, John Ky wrote:
Hi Haskell Cafe,
I tried using type families over functions, but when I try it complains that the two lines marked conflict with each other.
class Broadcast a where type Return a broadcast :: a -> Return a
instance Broadcast [a -> r] where type Return [a -> r] = a -> [r] -- Conflict! broadcast fs a = []
instance Broadcast [a -> b -> r] where type Return [a -> b -> r] = a -> b -> [r] -- Conflict! broadcast fs a b = []
Given that in Haskell, every function of n+1 arguments is also a function of n arguments, this is likely the cause of the conflict.
This solution is somewhat in-extensible in the ultimate result type (r, in your code); if the number of types r can take on is limited, it should work well, though. For expository purposes, I assume that r is always Int:
-- | Conal Elliot's semantic editor combinator argument argument :: (a -> a') -> (a -> b) -> (a' -> b) argument f g = g . f
class Broadcast a where type Return a broadcast :: [a] -> Return a instance Broadcast Int where type Return Int = [Int] broadcast ns = ns instance Broadcast r => Broadcast (a -> r) where type Return (a -> r) = a -> Return r broadcast fs x = (map.argument) (const x) fs
jcc