[Haskell-cafe] Re: Monads from Functors

[David Menendez]

On Wed, Apr 8, 2009 at 5:20 PM, Ben Franksen <ben.franksen at online.de> wrote:
> 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 <dave at zednenem.com>
<http://www.eyrie.org/~zednenem/>