[Haskell-cafe] Monads from Functors
Edward Kmett
ekmett at gmail.com
Wed Apr 8 17:23:33 EDT 2009
I discovered them and bundled them up a year or so back in category-extras.
http://comonad.com/haskell/category-extras/dist/doc/html/category-extras/Control-Monad-Codensity.html
I also wrote a series of blog posts including the derivation of these and
their dual in the form of right- and left- Kan extensions.
http://comonad.com/reader/2008/kan-extensions/
http://comonad.com/reader/2008/kan-extensions-ii/
http://comonad.com/reader/2008/kan-extension-iii/
I shared with Janis Voigtlaender the connection to his asymptotic
improvement in the performance of free monads paper as well. After I
discovered the connection between these and that paper shortly thereafter.
-Edward Kmett
On Wed, Apr 8, 2009 at 2:22 PM, Sebastian Fischer <
sebf at informatik.uni-kiel.de> wrote:
> > {-# LANGUAGE Rank2Types #-}
>
> Dear Haskellers,
>
> I just realized that we get instances of `Monad` from pointed functors
> and instances of `MonadPlus` from alternative functors.
>
> Is this folklore?
>
> > import Control.Monad
> > import Control.Applicative
>
> In fact, every unary type constructor gives rise to a monad by the
> continuation monad transformer.
>
> > newtype ContT t a = ContT { unContT :: forall r . (a -> t r) -> t r }
> >
> > instance Monad (ContT t)
> > where
> > return x = ContT ($x)
> > m >>= f = ContT (\k -> unContT m (\x -> unContT (f x) k))
>
> Both the `mtl` package and the `transformers` package use the same
> `Monad` instance for their `ContT` type but require `t` to be an
> instance of `Monad`. Why? [^1]
>
> If `f` is an applicative functor (in fact, a pointed functor is
> enough), then we can translate monadic actions back to the original
> type.
>
> > runContT :: Applicative f => ContT f a -> f a
> > runContT m = unContT m pure
>
> If `f` is an alternative functor, then `ContT f` is a `MonadPlus`.
>
> > instance Alternative f => MonadPlus (ContT f)
> > where
> > mzero = ContT (const empty)
> > a `mplus` b = ContT (\k -> unContT a k <|> unContT b k)
>
> That is no surprise because `empty` and `<|>` are just renamings for
> `mzero` and `mplus` (or the other way round). The missing piece was
> `>>=` which is provided by `ContT` for free.
>
> Are these instances defined somewhere?
>
> Cheers,
> Sebastian
>
> [^1] I recognized that Janis Voigtlaender defines the type `ContT`
> under the name `C` in Section 3 of his paper on "Asymptotic
> Improvement of Computations over Free Monads" (available at
> http://wwwtcs.inf.tu-dresden.de/~voigt/mpc08.pdf) and gives a monad
> instance without constraints on the first parameter.
>
> _______________________________________________
> Haskell-Cafe mailing list
> Haskell-Cafe at haskell.org
> http://www.haskell.org/mailman/listinfo/haskell-cafe
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: http://www.haskell.org/pipermail/haskell-cafe/attachments/20090408/2e419499/attachment.htm
More information about the Haskell-Cafe
mailing list