Reputation: 27217
The following code is intended to produce either a Double or an Integer. s
is assumed to be either negate
or id
; n
the whole part; and f
the fractional part or Nothing
for an integer.
computeValue :: Num a => (a->a) -> Integer -> (Maybe Double) -> Either Double Integer
computeValue s n Nothing = Right $ s n
computeValue s n (Just a) = Left $ s (fromIntegral n + a)
when I compile this I get:
test1.hs:2:28:
Couldn't match type `Integer' with `Double'
Expected type: Either Double Integer
Actual type: Either Double a
In the expression: Right $ s n
In an equation for `computeValue':
computeValue s n Nothing = Right $ s n
test1.hs:2:38:
Couldn't match type `Integer' with `Double'
In the first argument of `s', namely `n'
In the second argument of `($)', namely `s n'
In the expression: Right $ s n
It seems like somehow the compiler has lost track of the fact that s
is polymorphic. What happened here and how do I fix it?
Upvotes: 7
Views: 365
Reputation: 120711
s
is not polymorphic from inside of your function: you can use any function that works on some Num
instance as this parameter, it might be a function that only works on Complex
! What you need is an universally quantified function s
, i.e. one that can actually be called with any Num
instance.
{-# LANGUAGE Rank2Types #-}
computeValue :: (forall a . Num a => a->a) -> Integer -> Maybe Double -> Either Double Integer
computeValue s n Nothing = Right $ s n
computeValue s n (Just a) = Left $ s (fromIntegral n + a)
That works then:
Prelude Data.Either> computeValue id 3 Nothing
Right 3
Prelude Data.Either> computeValue negate 57 (Just pi)
Left (-60.1415926535898)
Upvotes: 10