On 4/27/06, Robin Bate Boerop
But, this code:
class CC a type C x = CC a => a x f, g :: C a -> Int f _ = 3 g x = f $ x -- the only change
The problem is exactly the use of $. $ is an operator, not a built-in language construct, and it has type (a -> b) -> a -> b. No forall's in there, so you cannot give it a function argument that is existentially quantified. Lots of people have been bitten by this when using the magic runST with type "forall a. (forall s. ST s a) -> a". Use parentheses when you have existentially quantified values and everything should be just fine. :-) /Niklas
gives this error:
Inferred type is less polymorphic than expected Quantified type variable `a' escapes Expected type: a a1 -> b Inferred type: C a1 -> Int In the first argument of `($)', namely `f' In the definition of `g': g x = f $ x
What's going on here?
-- Robin Bate Boerop
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