[Haskell-cafe] Existentially-quantified constructors, Eq and Show
joelr1 at gmail.com
Wed Dec 7 17:12:07 EST 2005
Is there a less verbose way of doing this:
data State a
| (Show a, Eq a) => State a
instance Eq a => Eq (State a) where
(State a) == (State b) = a == b
Start == Start = True
Stop == Stop = True
instance Show a => Show (State a) where
show (State a) = show a
show Start = "Start"
show Stop = "Stop"
More information about the Haskell-Cafe