Re: [Haskell-cafe] Re: Applicative but not Monad

On Sat, Oct 31, 2009 at 8:38 AM, David Menendez wrote:

On Sat, Oct 31, 2009 at 6:22 AM, Heinrich Apfelmus wrote:

...

The only possible monad instance would be

return x = Const mempty fmap f (Const b) = Const b join (Const b) = Const b

but that's not just () turned into a monad.

This is inconsistent with the Applicative instance given above.

Const a <*> Const b = Const (a `mappend` b)

Const a `ap` Const b = Const a

In other words, Const has at least two Applicative instances, one of which is not also a monad.

But this "Monad" instance isn't a monad either: f True = Const [1] f False = Const [2] return True >>= f {- by monad laws -} = f True = Const [1] but by this code return True >>= f {- apply return, monoid [a] -} = Const [] >>= f {- definition of >>= -} = join (fmap f (Const [])) {- apply join and fmap -} = Const []