On Wed, Apr 8, 2009 at 5:20 PM, Ben Franksen
Sebastian Fischer 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]
Maybe because this is needed to prove monad laws?
<snip derivation>
So, that wasn't the reason. It really is a monad.
In general, ContT r m a is equivalent to Cont (m r) a, and their
corresponding Monad instances are also equivalent. But Cont r is a
monad for any r, which implies that ContT r m must be a monad for any
r and m.
--
Dave Menendez