But by doing so I am changing type equality to the same as having "type
family F a :: * -> *"
F' a x ~ F' b y <=> F' a ~ F' b /\ x ~ y (equality for ADTs)
and I would like this "decomposition rule not to apply" so.
Thanks though,
hugo
On Tue, Mar 18, 2008 at 1:12 AM, Ryan Ingram <[EMAIL PROTEC
On 3/17/08, Hugo Pacheco <[EMAIL PROTECTED]> wrote:
> type family G a :: * -> *
> type instance G Int = Either () -- you forgot this line
>
> instance Functor (G Int) where
>fmap f (Left ()) = Left ()
>fmap f (Right x) = Right (f x)
One thing that you don't seem to be clear about is that t
Hi,
I am trying to understand some differences of parameterizing or not some
arguments of type families.
I have some code such as
*type family G a :: * -> *
instance Functor (G Int) where
fmap f (Left ()) = Left ()
fmap f (Right x) = Right (f x)
ggg :: Functor (G a) => G a x -> G a x
ggg
