Why is this type variable ambiguous?
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23-08-2019 - |
Question
Cabbage.hs:
module Cabbage where
class Cabbage a
where foo :: a -> String -- the parameter is only present for its type,
-- the parameter value will be ignored
bar :: String -> a
quux :: Cabbage a => String -> a
quux s = bar (s ++ foo (undefined :: a))
When I compile (with ghc) I get this error message:
Cabbage.hs:7:19:
Ambiguous type variable `a' in the constraint:
`Cabbage a' arising from a use of `foo' at Cabbage.hs:7:19-38
Probable fix: add a type signature that fixes these type variable(s)
I don't understand why a
is ambiguous. Surely the a
in line 7 is the same as the a
in line 6? How do I fix this?
Alternatively, is there a better way of declaring a per-instance constant?
Solution
Using scoped type variables you can let GHC know that the undefined :: a
should be the same (otherwise a
is just a shorthand for forall a. a
). Scoped type variables must then be explicitly forall-qualified:
{-# LANGUAGE ScopedTypeVariables #-}
module Cabbage where
class Cabbage a
where foo :: a -> String -- the parameter is only present for its type,
-- the parameter value will be ignored
bar :: String -> a
quux :: forall a. Cabbage a => String -> a
quux s = bar (s ++ foo (undefined :: a))
OTHER TIPS
The problem is that Haskell doesn't know which instance of Cabbage
that foo
corresponds to there. So far as I know, it doesn't match the a
in (undefined :: a)
with the a
in quux :: Cabbage a => String -> a
Assuming that's what you want, you can do this:
quux :: Cabbage a => String -> a
quux s = result
where result = bar (s ++ foo result)
This ties foo and bar together so that it uses the same instance for both, and since you don't actually need the value of the input for foo, it bottoms out. I don't know of a better way of doing per-instance constants though. Hopefully someone else will come along who does.
you can extract out the polymorphic part as a function
quux :: Cabbage a => String -> a
quux s = quux' undefined
where quux' :: Cabbage a => a -> a
quux' x = bar (s ++ foo x)