Re: Cont as Monoid

"Cont (Endo b) a" is the usual backtracking monad.

It is? Would you say more about that? A pointer would be fine. I'm wondering what the role of Endo is here. Cheers, - Conal On 9/9/07, David Menendez wrote:

On 9/8/07, Conal Elliott wrote:

...

I'd like to see the following addition to Control.Monad.Cont in mtl:

instance Monoid r => Monoid (Cont r a) where mempty = Cont mempty m `mappend` m' = Cont (runCont m `mappend` runCont m')

Alternatively,

instance Monoid r => MonadPlus (Cont r a) where mzero = Cont mempty mplus a b = Cont $ runCont a `mappend` runCont b

"Cont (Endo b) a" is the usual backtracking monad.