New instance declaration for Show

To derive an instance, the syntax is

instance «preconditions» => Class «type» where
  «method» = «definition»

So here, for instance, you’d have

instance Show Prediction where
  show (Prediction a b c) = show a ++ "-" ++ show b ++ "-" ++ show c

There’s no precondition; you’d use that for something like instance Show a => Show [a] where ..., which says that if a is showable, then so is [a]. Here, all Predictions are showable, so there’s nothing to worry about. When you wrote instance Show (Prediction p) => showPrediction p, you made a few mistakes. First, Prediction p implies that Prediction is a parametrized type (one declared by, for instance, data Prediction a = Prediction a a a), which it isn’t. Second, Show (Prediction p) => implies that if Prediction P is showable, then you want to declare some other instance. And third, after the =>, having a function is nonsensical—Haskell wanted a type class name.

Also, for completeness’s sake, there’s another way to derive Show if you want the Prediction 1 2 3 format for displayed output:

data Prediction = Prediction Int Int Int deriving Show

As specified in the Haskell 98 report, there are only a handful of types which can be derived this way: Eq, Ord, Enum, Bounded, Show, and Read. With the appropriate GHC extensions, you can also derive Data, Typeable, Functor, Foldable, and Traversable; you can derive any class which a newtype‘s wrapped type derived for a newtype; and you can generate these automatic instances in a standalone way.

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