Simon Peyton-Jones wrote:
I think you can simplify the example. Given
class HasFoo a b | a -> b where foo :: a -> b
instance HasFoo Int Bool where ...
Is this legal?
f :: HasFoo Int b => Int -> b f x = foo x
You might think so, since HasFoo Int b => Int -> b is a substitution instance of HasFoo a b => a -> b
This is the step where the reasoning goes wrong. The functional dependency tells you that `b' isn't really a free variable, since it is dependent on `a'. If you substitute for `a', you can't expect `b' to remain unconstrained. Hugs complains that the inferred type for `f' is not general enough. It's right to complain, but the real problem is that the signature is too general. Asimilar situation arises if you try to declare an instance `HasFoo Int b', but in this case, hugs complains that the instance is more general than the dependency allows. A useful thing to do would be to check for this sort of thing in signatures as well, so that the more appropriate error message can be given. --Jeff