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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