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

I glossed over the key fact

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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 wrote:

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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.

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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?

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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?

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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)

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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

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There are similar monoidal instances for Bool, such as Any and All.

On Wed, Jun 7, 2017 at 12:33 PM, Baa wrote:

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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 _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

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-- Best regards, Paul a.k.a. 6apcyk _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

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-- Best regards, Paul a.k.a. 6apcyk