Want an "f a -> f b"?
Got a->b? Use <$>
Got f (a->b)? Use <*>
Got a -> f b? Use =<<
#haskell
-
-
@dibblego@bitemyapp@puffnfresh Sorry, was unclear, it's not. Just trying to make sure I understand. -
@JulianBirch@bitemyapp@puffnfresh Provide an identity to extend that has the type Func<IObservable<A>, A> - Show replies
New conversation -
Loading seems to be taking a while.
Twitter may be over capacity or experiencing a momentary hiccup. Try again or visit Twitter Status for more information.