[Haskell-beginners] mempty and "No instance for (Monoid Int)"

Wed Jun 7 20:05:06 UTC 2017

> I glossed over the key fact
> 
> > Maybe Int is only a Monoid if Int is an instance of Monoid  
> 
> This is derived from the Monoid instance of Maybe.
> 
> instance Monoid a => Monoid (Maybe a) -- Defined in ‘GHC.Base’

Ooohh... I see. Thank you a lot!
No magic again :)

/Best regards,
  Paul

> 
> Maybe is only an instance if a is an instance.  If a isn't then Maybe
> isn't either, and it will be rejected.  That is why Maybe Int is not a
> Monoid, but Maybe (Product Int) and Maybe () are.
> 
> 
> On Wed, Jun 7, 2017 at 2:53 PM, aquagnu <aquagnu at gmail.com> wrote:
> >> In ghci there are type defaulting rules.  When you go mempty ==
> >> Nothing, it type defaults it to "Maybe ()".  
> >
> > Aha, OK. These defaults are preset also in non-interactive: I tried
> > the same and get the same result.
> >  
> >> But when you type Just 4,
> >> 4 is definitely not (), and so it looks at the Monoid instance for
> >> the a default type to use in cases of numeric literals, the first
> >> of which is Int.  
> >
> > This I can not understand. Literal "4" is under "Just", so why are
> > we talking about "Int" as Monoid but not about "Maybe Int" as
> > Monoid? And "Maybe Int" as Monoid does not depend on Int and is the
> > same for "Maybe Int", "Maybe Bool", "Maybe String"... When I added
> > type annotation, like "::Maybe Int", I suppose, usual "Maybe a"'s
> > implementations of "mempty", "mappend" will be used, - no more
> > defaults. Seems it is not true, but why?
> >
> >  
> >>
> >> Which brings you to the next problem.  Maybe Int is only a Monoid
> >> if Int is an instance of Monoid, and Int is definitely not.
> >>  
> >
> > I don't understand it. Monoid is "Maybe a" for any "a". And I can
> > understand your point if we are talking only for interactive GHCI
> > and its defaults, but when I tried in source code to write:
> >
> >   m :: Maybe Int
> >   m = mempty
> >   ...
> >   ... print $ Nothing == m
> >
> > i get the same, about no instance for (Monoid Int). But Maybe's
> > "mempty" is "Nothing", nothing else. And its "mappend" processes
> > any (Just _) and Nothing's, right? May be all magic is from
> > defaults?
> >
> >  
> >> That's because is 3 `mappend` 3 == 6 via addition?  Or should it
> >> be 9 via multiplication?  Or something else?  What should mempty
> >> be, 0?  Or maybe 1?  Who is to decide what the only way of
> >> combining Ints together is.
> >>
> >> It turns out there are instances for both of those cases, but you
> >> have to wrap the int into a type so that it knows which way you
> >> want it to be interpreted.
> >>
> >> import Data.Monoid
> >> mempty == Just (Product 1)  
> >> > false  
> >> mempty == Just (Sum 1)  
> >> > false  
> >
> > Yes, this is absolutely understandable. Except one detail:
> >
> >   Prelude Data.Monoid Data.Maybe> mempty == Product 1
> >   True
> >   Prelude Data.Monoid Data.Maybe> mempty == Just (Product 1)
> >   False
> >
> > so, "Product Int" as Monoid and "Maybe (Product Int)" as Monoid are
> > totally different, - I understand what is Abel's groups on + and *,
> > but I don't understand why GHC looks for Monoid instance for Int
> > while Int is under Maybe... It will be right if:
> >
> >   instance (Monoid a) => Monoid (Maybe a) where
> >     ...
> >
> > but is it true?! I suppose no such constraint on "a". Is it all due
> > to defaults? Or I lost my brain at this night :)
> >
> >
> > /Best regards, Paul
> >
> >  
> >>
> >> There are similar monoidal instances for Bool, such as Any and All.
> >>
> >> On Wed, Jun 7, 2017 at 12:33 PM, Baa <aquagnu at gmail.com> wrote:  
> >> > Maybe a is the Monoid:
> >> >
> >> >   instance Monoid a => Monoid (Maybe a) -- Defined in ‘GHC.Base’
> >> >
> >> > so I can compare its values with empty value:
> >> >
> >> >   mempty == Nothing  
> >> >   => True  
> >> >
> >> > But if I try:
> >> >
> >> >   mempty == Just 4
> >> >
> >> > I get:
> >> >
> >> >   <interactive>:1:1: error:
> >> >       • Ambiguous type variable ‘a0’ arising from a use of
> >> > ‘mempty’ prevents the constraint ‘(Monoid a0)’ from being solved.
> >> >         Probable fix: use a type annotation to specify what ‘a0’
> >> > should be. These potential instances exist:
> >> >           instance Monoid a => Monoid (IO a) -- Defined in
> >> > ‘GHC.Base’ instance Monoid Ordering -- Defined in ‘GHC.Base’
> >> >           instance Monoid a => Monoid (Maybe a) -- Defined in
> >> > ‘GHC.Base’ ...plus 7 others
> >> >           (use -fprint-potential-instances to see them all)
> >> >       • In the first argument of ‘(==)’, namely ‘mempty’
> >> >         In the expression: mempty == Just 4
> >> >         In an equation for ‘it’: it = mempty == Just 4
> >> >
> >> > OK, I try:
> >> >
> >> >   mempty::Maybe Int
> >> >
> >> > and get:
> >> >
> >> >   <interactive>:1:1: error:
> >> >       • No instance for (Monoid Int) arising from a use of
> >> > ‘mempty’ • In the expression: mempty :: Maybe Int
> >> >         In an equation for ‘it’: it = mempty :: Maybe Int
> >> >
> >> > so, how is related Int to Monoid, why does ghc expect from
> >> > mempty::Maybe Int, Int to be Monoid?! As I understand, this means
> >> > only that I mean "mempty" from (Maybe Int) type, which is Monoid
> >> > and exists sure.
> >> >
> >> > Interesting is, that:
> >> >
> >> >   mempty::Maybe [Int]  
> >> >   => Nothing  
> >> >
> >> > but how is related "monoidality" of "Maybe a" with "monoidality
> >> > of "a" ???
> >> >
> >> > Initial idea was to make comparison:
> >> >
> >> >   mempty :: Maybe Int == Just 4  
> >> >   => False  
> >> >
> >> >
> >> > /Best regards,
> >> >   Paul
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> >
> > --
> > Best regards,
> >   Paul a.k.a. 6apcyk
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-- 
Best regards,
  Paul a.k.a. 6apcyk