Janis Voigtlaender <voigt <at> tcs.inf.tu-dresden.de> writes: > > DavidA wrote: > > I suspect that the answer to the question is, yes, you can have different > > Functor instances. All you need is a sum-product type that it's possible to > > interpret as two different abstractions, leading to two different Functor > > instances. > > The sum-product types are exactly the "not-too-exotic" types to which my > proof applies. So as long as extensional equivalence means Haskell > equivalence, and not some "modulo an interpretation" equivalence (like > considering two lists equivalent if they contain the same elements but > in potentially different order), the answer is no, one cannot have > different funtor instances. > > Ciao, Janis. >
Okay, I see. Well that's interesting, because it suggests that your proof might break under modest extensions to the language. The thing that led me to suggest list / set as an example was consideration of Data.Set. Conceptually, this should be a Functor instance - given f :: a -> b, we should be able to derive fmap f :: Data.Set a -> Data.Set b, eg fmap f xs = (fromList . map f . toList) xs However, we can't currently express this in Haskell, because Data.Set requires Ord a. However, there have been some proposals to fix this. I assume that if they went through, then it would be possible to define a type which could be interpreted as either a list or a set, with different functor instances in either case. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
