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
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