How can I make (a, a) a Functor without resorting to a newtype?
Basically I want it to work like this:
instance Functor (a, a) where
fmap f (x, y) = (f x, f y)
But of course that's not a legal way to express it:
Kind mis-match
The first argument of `Functor' should have kind `* -> *',
but `(a, a)' has kind `*'
In the instance declaration for `Functor (a, a)'
What I really want is a type-level function like this: \a -> (a, a) (invalid syntax). So a type alias, perhaps?
type V2 a = (a, a)
instance Functor V2 where
fmap f (x, y) = (f x, f y)
I would think this would work, but it doesn't. First I get this complaint:
Illegal instance declaration for `Functor V2'
(All instance types must be of the form (T t1 ... tn)
where T is not a synonym.
Use -XTypeSynonymInstances if you want to disable this.)
In the instance declaration for `Functor V2'
If I follow the advice and add the TypeSynonymInstances extension, I get a new error:
Type synonym `V2' should have 1 argument, but has been given 0
In the instance declaration for `Functor V2'
Well, duh, that's the point! V2 has kind * -> * which is what is required of a Functor instance. Well, ok, I can use a newtype like this:
newtype V2 a = V2 (a, a)
instance Functor V2 where
fmap f (V2 (x, y)) = V2 (f x, f y)
But now I've got to sprinkle V2s liberally throughout my code instead of just being able to deal with simple tuples, which kind of defeats the point of making it a Functor; at that point I might as well make my own function vmap :: (a -> b) -> (a, a) -> (b, b).
So is there any way to do this nicely, i.e. without a newty