Hi,

what you are trying to accomplish is to derive a less specific instance from a more specific one (e.g. trying to derive Eq a, when you have Ord a). While this might seem natural from a mathematical point of view, in Haskell it is not. There are several possible solutions, one of those is given by the error message itself (turn on the language extension FlexibleInstances). Different instances of this general problem are discussed in [1], [2], [3].

Now as for your problem itself: You define a new type class, which already requires an Eq constraint. Then you try to derive an Eq context for certain types from the fact that they belong to the type class ThatsMyProblem. The context of ThatsMyProblem already provides an Eq context by definition and conceptually it should not be possible to define (==) in terms of "fromMyProblem", because the concept of "fromMyProblem" requires a definition of (==).

Regards,

Nikita

[1] http://stackoverflow.com/questions/8633470/illegal-instance-declaration-when-declaring-instance-of-isstring

[2] http://stackoverflow.com/questions/4826630/type-class-problem-concerning-flexibleinstances

[3] http://connectionrequired.com/blog/2009/07/my-first-introduction-to-haskell-extensions-flexibleinstances/

On 02/08/13 13:32, Mateusz Neumann wrote:
Hi

I came across a problem, which I deeply believe, might be solved in
Haskell in much nicer way (than I did it).  I have:

class (Eq a) => ThatsMyProblem a where
   fromMyProblem :: a -> Int
   toMyProblem :: Int -> a

data MyType1
    = MyType1_1
    | MyType1_2
    | MyType1_3 Int
  deriving (Show)
instance Eq MyType1 where
    (==) a b = fromMyProblem a == fromMyProblem b
instance ThatsMyProblem MyType1 where
    [...]

data MyType2
    = MyType2_1
    | MyType2_2 Int
  deriving (Show)
instance Eq MyType2 where
    (==) a b = fromMyProblem a == fromMyProblem b
instance ThatsMyProblem MyType2 where
    [...]

data MyType3
    = MyType3_1
    | MyType3_2
    | MyType3_3 Int
  deriving (Show)
instance Eq MyType3 where
    (==) a b = fromMyProblem a == fromMyProblem b
instance ThatsMyProblem MyType3 where
    [...]


I would very much like to create one single instance like this:

instance (FutureVal a) => Eq a where
    (==) x y = fromFVal x == fromFVal y

but that does not seem to work, as I get an error stating “Illegal
instance declaration for `Eq a' (All instance types must be of the form
(T a1 ... an) where a1 ... an are *distinct type variables*, and each
type variable appears at most once in the instance head. Use
-XFlexibleInstances if you want to disable this.) In the instance
declaration for `Eq a'”

Could you please point me out my mistake and/or direct me to some
documentation?




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