6 May
2017
6 May
'17
6:25 p.m.
mappend pb1 pb2 = FunPB $ \n -> (,) n $ msum . (<$>) (uncurry (flip const).((flip runFunPB) n)) $ [pb1,pb2]
This cracks me up. Well played. Well played indeed. Well all the things you need are right here in this line, right? I mean you sent it because you want to show us what you know, right? There's higher order functions, there's mapping and folding and currying of different kinds and everything. So what's missing?