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

I was also hoping that something like this would let me avoid quantifying a in functions downstream but alas, it does not happen. I have to use (Eq a, Show a) => a ... everywhere else. On Dec 7, 2005, at 11:14 PM, Bulat Ziganshin wrote:

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

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