On Sat, Oct 31, 2009 at 6:22 AM, Heinrich Apfelmus
<apfelmus@quantentunnel.de> 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