How do I use a monoid instance of a function?

Question

Today I tried to reduce a list of functions trough monoid typeclass but the resulting function expects its argument to be an instance of Monoid for some reason.

GHCI tells me that the type of mconcat [id, id, id, id] is Monoid a => a -> a. Yet I would expect it to be a -> a.

What is happening?

score 12 · Accepted Answer · answered Jun 03 '17 at 14:13

12

You're using this instance:

instance Monoid b => Monoid (a -> b) where
    mempty _ = mempty
    mappend f g x = f x `mappend` g x

which is more general because it doesn't require endomorphisms (i.e. a -> a). To get the instance you were expecting, you can wrap your functions in Endo:

appEndo (mconcat [Endo id, Endo id, Endo id, Endo id])

or

appEndo $ mconcat $ fmap Endo [id, id, id, id]

answered Jun 03 '17 at 14:13

user2297560

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Yet another alternative: `appEndo $ foldMap Endo [id,id,id,id]` – chi Jun 03 '17 at 14:44
So, `Monoid Endo` has `mempty = id ; mappend = (.)`, if I understand correctly? – schuelermine Jun 24 '18 at 08:05

How do I use a monoid instance of a function?

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