> {-# 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.
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