C T McBride wrote:
Hi
This is a long message, containing a program which makes heavy use of type classes with functional dependencies, and a query about how the typechecker treats them. It might be a bit of an effort, but I'd be grateful for any comment and advice more experienced Haskellers can spare the time to give me. Er, this ain't no undergrad homework...
Jeffrey R. Lewis:
Without delving too deeply into your example, it looks like you've bumped into a known bug in Hugs implementation of functional dependencies. You should try GHCI if you can - it doesn't suffer from this bug.
Thanks for the tip! Our local Haskell supremo has pointed me to a version I can run, and that has improved the situation... a bit. Now I get Perm> permArg (FS (S O) (FO O),(FO O,())) No instance for `Show ((r -> s -> s) -> s -> r -> s)' Obviously, there's no Show method, but I was expecting a more general type. But if I tell it the right answer, it believes me Perm> permArg (FS (S O) (FO O),(FO O,())) :: (a -> b -> c) -> b -> a -> c No instance for `Show ((a -> b -> t) -> b -> a -> t)' The main thing is that I can permute a function correctly, without needing an explicit signature: Perm> permArg (FS (S (S O)) (FO (S O)),(FS (S O) (FO O),(FO O,()))) foldl No instance for `Show ([b] -> (t -> b -> t) -> t -> t)' Still, those too-specific inferred types are disturbing me a little Perm> permArg (FS (S (S O)) (FO (S O)),(FS (S O) (FO O),(FO O,()))) No instance for `Show ((r -> s -> s -> s) -> s -> r -> s -> s)' The above examples show that my code works with the generality it needs to. Is there some `defaulting' mechanism at work here for the inferred types making all those s's the same? But this is definitely progress! Thanks Conor