Prelude> :t id :: Eq b => b -> b id :: Eq b => b -> b :: (Eq b) => b -> b Prelude> id :: Eq b => b -> b
<interactive>:1:0: No instance for (Show (b -> b)) arising from a use of `print' at <interactive>:1:0-19 Possible fix: add an instance declaration for (Show (b -> b)) In a stmt of a 'do' expression: print it
The Eq constraint is irrelevant to the fact that there is no b -> b Show instance. The search for an instance looks only at the type part, and if it finds a suitable match, then checks if the necessary constraints are also in place.
Thanks - that first sentence is right, but not because of the second!-) I thought the error message was misleading, speaking of instance when it should have been speaking about instance *head*. But I was really confused by the difference between (making the show explicit and using scoped type variables): *Main> :t (show (id::a->a))::forall a.Eq a=>String <interactive>:1:0: Ambiguous constraint `Eq a' At least one of the forall'd type variables mentioned by the constraint must be reachable from the type after the '=>' In an expression type signature: forall a. (Eq a) => String In the expression: (show (id :: a -> a)) :: forall a. (Eq a) => String <interactive>:1:1: Could not deduce (Show (a -> a)) from the context (Eq a) arising from a use of `show' at <interactive>:1:1-15 Possible fix: add (Show (a -> a)) to the context of an expression type signature or add an instance declaration for (Show (a -> a)) In the expression: (show (id :: a -> a)) :: forall a. (Eq a) => String which does list a non-empty context from which 'Show (a->a)' could not be deduced and *Main> :t show (id :: Eq a => a-> a) <interactive>:1:0: No instance for (Show (a -> a)) arising from a use of `show' at <interactive>:1:0-25 Possible fix: add an instance declaration for (Show (a -> a)) In the expression: show (id :: (Eq a) => a -> a) which only talks about general 'Show (a->a)' instances. We cannot define 'instance Show (forall a. Eq a=>a->a)', which is why the 'Eq a' constraint plays no role, I think.. Btw, the issues in the other example can be made more explicit by defining: class MyEq a b | a->b, b->a instance MyEq a a First, both FDs and TFs simplify this: *Main> :t id :: (forall b. MyEq b Bool => b->b) id :: (forall b. MyEq b Bool => b->b) :: Bool -> Bool *Main> :t id :: (forall b. b~Bool => b->b) id :: (forall b. b~Bool => b->b) :: Bool -> Bool but the FD version here typechecks (note, though, that the type is only partially simplified) *Main> :t id :: (forall b. MyEq b Bool => b->b) -> (forall b. MyEq b Bool => b->b) id :: (forall b. MyEq b Bool => b->b) -> (forall b. MyEq b Bool => b->b) :: (forall b. (MyEq b Bool) => b -> b) -> Bool -> Bool while the TF version doesn't *Main> :t id :: (forall b. b~Bool => b->b) -> (forall b. b~Bool => b->b) <interactive>:1:0: Couldn't match expected type `forall b. (b ~ Bool) => b -> b' against inferred type `forall b. (b ~ Bool) => b -> b' Expected type: forall b. (b ~ Bool) => b -> b Inferred type: forall b. (b ~ Bool) => b -> b In the expression: id :: (forall b. (b ~ Bool) => b -> b) -> (forall b. (b ~ Bool) => b -> b) Claus