[Haskell-beginners] Functor question.

[Alexander Dunlap]

On Sun, Nov 15, 2009 at 7:50 PM, Phillip Pirrip <ppirrip at gmail.com> wrote:
>
> Hi,
>
> When I define my data as fellow,
>
> data Moo a = Moo a
>            deriving (Show)
>
> instance Functor Moo where
>   fmap f (Moo a) = Moo (f a)
>
> GHC gives me no problem.  Then I add something,
>
> data (Num a) => Moo a = Moo a
>            deriving (Show)
>
> instance Functor Moo where
>   fmap f (Moo a) = Moo (f a)
>
> Now GHC gives me the follow error - see bellow.
>
> What is the reason behind this?  What should I do to correct this?
>
> thx,
>
> //
>
> matFun_v2.hs:16:12:
>   Could not deduce (Num a) from the context ()
>     arising from a use of `Moo' at matFun_v2.hs:16:12-14
>   Possible fix:
>     add (Num a) to the context of the type signature for `fmap'
>   In the pattern: Moo a
>   In the definition of `fmap': fmap f (Moo a) = Moo (f a)
>   In the instance declaration for `Functor Moo'
>
> matFun_v2.hs:16:21:
>   Could not deduce (Num b) from the context ()
>     arising from a use of `Moo' at matFun_v2.hs:16:21-29
>   Possible fix:
>     add (Num b) to the context of the type signature for `fmap'
>   In the expression: Moo (f a)
>   In the definition of `fmap': fmap f (Moo a) = Moo (f a)
>   In the instance declaration for `Functor Moo'
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Your datatype Moo cannot be an instance of Functor because Moo cannot
contain all types. fmap takes a function of type (a -> b), which means
that the function can be from *any* type a to *any* type b, and
produces a function of type (f a -> f b), which in your case means Moo
a -> Moo b. But the function fmap f (Moo x) = Moo (f x) does not have
type "a -> b"; it has type "(Num a, Num b) => a -> b", since if you
can have a type Moo a, a must be an instance of Num.

In general, people recommend against using constriants on datatypes,
recommending instead to put those constraints on the functions that
operate on the datatypes. I'm not quite sure why that is, though.

Alex