instance forall m hg. (HasResolver m hg, Functor m, ToValue (Maybe hg)) => HasResolver m (Maybe hg) where
type Handler m (Maybe hg) = m (Maybe hg)
resolve handler _ = map (ok . toValue) handler
Note that it discards the selection set. That can't be good.
Nope. Removing it broke the tests. Just pushed a revised #101 which restores the code together with a comment that explains what it does and why it's broken.
Probably.
Code is:
Note that it discards the selection set. That can't be good.