Adrian Hey wrote:
oleg@pobox.com wrote:
Adrian Hey wrote:
-- Instances of GT are instances of Eq -- instance (GT map key, Eq a) => Eq (map a) where map1 == map2 = assocsAscending map1 == assocsAscending map2 ... Overlapping instances for Eq [(key, a)] arising from use of `==' at Test.hs:10:16-59 Matching instances: instance (Eq a) => Eq [a] -- Defined in GHC.Base instance (GT map key, Eq a) => Eq (map a) -- Defined at Test.hs:9:0
How can my new instance overlap with the old (ghc) instance unless [] is also an instance of GT for some key type (which it isn't). Could someone explain?
You are right in your explanation of the GHC behavior. Your instance |Eq (map a)| indeed overlaps the `standard` instance |Eq [a]|. The overlap may be easier to see if we write [a] as ([] a), which is what it is, an application of the type constructor [] to the type variable a. So, the type [a] (aka [] a) is a particular instance of the type (map a), with `map' being []. This is the overlapping that GHC is complaining about.
So if I understand correctly, it's complaining about a potential overlap, not an actual overlap (there is no actual overlap until [] is made an instance of GT AFAICS).
Also, I suspect I'm still missing something important here, for example I don't understand why, if it overlaps for [], it doesn't overlap with other instances (like Maybe for example). Or am I just not getting the error for Maybe because ghc stops after the first error?
Ah, this brings me to something else I don't quite understand about this error, but kind of explains the absence of similar errors for Maybe. It's the "arising from use of `==' at Test.hs:10:16-59" part. I don't understand why, but if I use.. -- Instances of GT are instances of Eq -- instance (GT map key, Eq a) => Eq (map a) where map1 == map2 = True ..the error goes away. So it's not objecting to such instances in general, but rather the specific use of == in the definition. So now I'm really confused. Why, if it's possible that this instance overlaps with another for [], does it make any difference whether or not I've used == in the definition? Surely, if there's any ambiguity about which instance of Eq to use for [] (should [] be made an instance of GT), then the ambiguity exists whether or not I've used == in the definition. Hmm.. Regards -- Adrian Hey