Am Donnerstag, 4. Dezember 2008 00:05 schrieb Anatoly Yakovenko:
module Test where --why does this work: data Test = Test
class Foo t where foo :: Num v => t -> v -> IO ()
instance Foo Test where foo _ 1 = print $ "one" foo _ _ = print $ "not one"
--but this doesn't?
class Bar t where bar :: Foo v => t -> v -> IO ()
instance Bar Test where bar _ Test = print $ "test" bar _ _ = print $ "not test"
Because bar has to work for all types which belong to class Foo, but actually uses the type Test. This is what the error message Test.hs:18:10: Couldn't match expected type `v' against inferred type `Test' `v' is a rigid type variable bound by the type signature for `bar' at Test.hs:15:15 In the pattern: Test In the definition of `bar': bar _ Test = print $ "test" In the definition for method `bar' tells you. In the signature of bar, you've said that bar works for all types v which are members of Foo. Test is a monomorphic value of type Test, so it can't have type v for all v which belong to Foo. It doesn't matter that there is so far only the one instance of Foo, there could be others defined in other modules. The first works because the type of 1 in the definition of foo is defaulted to Integer (or whatever you specified in the default declaration).