It can be even worse:
data X a b = X (X b a -> b)
Here (X a) is certainly a functor, but the implementation must also
act on "a" contravariantly:
mapX :: (a -> a') -> X a' b -> X a b
mapX f (X h) = X $ h . fmap f
instance Functor (X a) where fmap f (X h) = X $ f . h . mapX f
On 5 Jun 2008, at 12:39, Thomas Davie wrote:
Even deriving an instance of Functor seems rather implausable, what
should it do for
data Wierd a b = Nil | A a (Wierd a b) | B b (Wierd a b)
Should fmap's function argument operate on 'a's, 'b's, or both?
Bob
On 5 Jun 2008, at 10:28, Miguel Mitrofanov wrote:
Well, it's certainly not possible for "filter", at least, not
without additional hints to the compiler. For example, consider
this type:
data Weird a = A | B a (Weird a) (Weird a)
filter p A = A
filter p (B x w1 w2) | p x = B x (filter p w1) (filter p w2)
| otherwise = ?????
On 5 Jun 2008, at 12:03, Cetin Sert wrote:
Hi ^_^,
Let's say we have the following data type and functions:
data Tab a = (:↺:)
| a :↓: Tab a
| Tab a :↙↘: (Tab a,Tab a)
deriving (Eq, Show, Read)
map f (:↺:) = (:↺:)
map f (a :↓: t) = f a :↓: map f t
map f (h :↙↘: (l,r)) = map f h :↙↘: (map f l, map f r)
filter p (:↺:) = (:↺:)
filter p (a :↓: t) | p a = filter p t
| otherwise = a :↓: filter p t
filter p (h :↙↘: (l,r)) = filter p h :↙↘: (filter p l,
filter p r)
is it possible to automatically derive map and filter?
data Tab a = (:↺:)
| a :↓: Tab a
| Tab a :↙↘: (Tab a,Tab a)
deriving (Eq, Show, Read, Map, Filter)
If not, do you think it might be nice to have something like this
in the future?
Best Regards,
Cetin Sert
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