[Haskell-cafe] Functor instance for FunPB

[Joachim Breitner]

Hi,

Am Samstag, den 06.05.2017, 20:32 +0200 schrieb Alex Wede:
> The monoid instance is from my teacher. 

it is hard to tell with confidence, given that we do not know what
FunFB is supposed to “mean”, but 

data FunPB a b = FunPB { runFunPB :: a -> (a,[b]) }
instance Monoid (FunPB a b) where
   mempty  = FunPB $ \k -> (k,mempty)
   mappend pb1 pb2 = FunPB $ \n -> (,) n $ msum . (<$>) (uncurry (flip const).((flip runFunPB) n)) $ [pb1,pb2]

looks very fishy, as the return value of of type a from pb1 and pb2 are ignored in mappend

The latter means that this is not a lawful monoid, because for

  pb = (\n -> (n+1, []) 

we have

  mempty `mappend` pb == mempty /= pb


Did you maybe mean

   mappend pb1 pb2 = FunPB  $ \n0 ->
	let (n1,xs1) = runFunFB n1
	    (n2,xs2) = runFunFB n2
        in (n2, xs1 ++ xs2)

Greetings,
Joachim


-- 
Joachim Breitner
  mail at joachim-breitner.de
  http://www.joachim-breitner.de/
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