Why is there not 'Alternative' instance for 'Control.Applicative.Const'
Asked Answered
E

1

16

There is an instance Monoid a => Monoid (Const a b) for the Const functor from Control.Applicative. There is also an instance Monoid m => Applicative (Const m).

I would therefore expect that there is also an instance Monoid m => Alternative (Const m) that coincides with the one for Monoid. Is this just an omission that should be fixed, or is there a deeper reason?

Ekg answered 23/2, 2015 at 18:56 Comment(1)
There is nothing deep about Alternative. Like MonadPlus, no one can even agree what its laws are supposed to be, and its most compelling use case is a sort of pun. Do whatever you want with it.Rockoon
A
8

I believe there is a deeper reason. While it seems there is no canonical set of rules for Alternative, in order for Alternative to make sense, there definitely ought to be a relationship between Alternative and its Applicative operations (otherwise it'd be just an arbitrary monoid.)

This answer to Confused by the meaning of the 'Alternative' type class and its relationship to other type classes states these laws

  1. Right distributivity (of <*>):  (f <|> g) <*> a = (f <*> a) <|> (g <*> a)
  2. Right absorption (for <*>):  empty <*> a = empty
  3. Left distributivity (of fmap):  f <$> (a <|> b) = (f <$> a) <|> (f <$> b)
  4. Left absorption (for fmap):  f <$> empty = empty

which make a lot of sense to me. Roughly speaking, empty and <|> are to pure and <$>/<*> what 0 and + are to 1 and *.

Now if we add instance Monoid m => Alternative (Const m) that coincides with the one for Monoid / Applicative, the Right- laws don't hold.

For example, 2. fails, because

empty <*> (Const x)
= Const mempty <*> Const x    -- by the suggested definition of Alternative
= Const $ mempty `mappend` x  -- by the definition of <*> for COnst
= Const x                     -- by monoid laws

which isn't equal to empty = Const mempty. Similarly, 1. fails, a simple counter-example is setting f = Const (Sum 1); g = Const (Sum 1) ; a = Const (Sum 1).

See also:

Alek answered 25/2, 2015 at 8:15 Comment(1)
Thank you very much for your detailed, well-written answer. I agree that these laws do make sense.Ekg

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