[Haskell-cafe] Existentially-quantified constructors, Eq and Show

[Joel Reymont]

Here is something else that I don't quite understand...

Original version compiles:

push :: Show b => State b -> Dispatcher b a -> (ScriptState a b) ()
push state dispatcher =
     do w <- get
        trace 95 $ "push: Pushing " ++ show state ++ " onto the stack"
        let s = stack w
        putStrict $ w { stack = (state, dispatcher):s }

data State a
     = Start
     | Stop
     | (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
     _ == _ = False

instance Show a => Show (State a) where
     show (State a) = show a
     show Start = "Start"
     show Stop = "Stop"

This version does not. Why does it require Eq in the ++ context? And  
why doesn't the other version?

data (Show a, Eq a) => State a
     = Start
     | Stop
     | State a
     deriving (Eq, Show)

Could not deduce (Eq b) from the context (Show b)
    arising from use of `show' at ./Script/Engine.hs:86:38-41
Probable fix: add (Eq b) to the type signature(s) for `push'
In the first argument of `(++)', namely `show state'
In the second argument of `(++)', namely `(show state) ++ " onto the  
stack"'

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