[Haskell-cafe] Functor instance for FunPB

Alex Wede awede at techfak.uni-bielefeld.de
Sat May 6 18:32:45 UTC 2017


No, I do not want to show anything.

The monoid instance is from my teacher. It is an exercise. Sry for the 



Am 06.05.2017 um 20:25 schrieb MarLinn:
>>   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?
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