Hallo. You can try reusing fmap.
data FunPB a b = FunPB { runFunPB :: a -> (a,[b]) fmap f (FunPB g) = FunPB $ fmap (fmap (fmap f)) g
This works because (a->), (a,), and [] are all functors.
Freundliche Grüße,
Daniel.
On Sat, May 6, 2017 at 9:04 PM, MarLinn
On 2017-05-06 20:38, David Kraeutmann wrote:
The only lawful instance I can think of is: instance Functor (FunPB a) where fmap f (FunPB g) = FunPB $ \k -> let (a,b) = g k in (a, fmap f b)
which is fairly straightforward IMO.
And because it's closer to the original style:
fmap f = FunPB .((.)((<$>).(<$>)$f)). runFunPB
Now it's STARRING INTO YOU SOUL. And it's also so much easier to read.
_______________________________________________ Haskell-Cafe mailing list To (un)subscribe, modify options or view archives go to: http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe Only members subscribed via the mailman list are allowed to post.