Move MonadIO to base

[David Menendez]

On Sun, Apr 18, 2010 at 5:02 PM, wren ng thornton
<wren at community.haskell.org> wrote:
> Heinrich Apfelmus wrote:
>>
>> Anders Kaseorg wrote:
>>>
>>> This concept can also be generalized to monad transformers:
>>>
>>> class MonadTrans t => MonadTransMorph t where
>>>    morph :: Monad m => (forall b. (t m a -> m b) -> m b) -> t m a
>>
>> [...]
>> However, not all control operators can be lifted this way. Essentially,
>> while you may "downgrade" an arbitrary selection of  t m a  values you
>> may only promote one  m a  in return and all have to share the same
>> return type  a . In particular, it's not possible to implement
>>
>>    lift :: (Monad m, MonadTrans t) => m a -> t m a
>
> Why not?
> * morph       says m(t m a) is a subset of (t m a)
> * Monad m     says we can fmap :: (a->b) -> (m a->m b)
> * Monad (t m) says we can return :: a -> t m a
>
>    lift ma = morph (\k -> k (fmap return ma))

Maybe something like this?

lift m = morph (\k -> m >>= k . return)
    -- n.b., return and >>= are from different monads

> Again, having m(t m a)->(t m a) is strictly more expressive than only having
> (m a)->(t m a) because the former may avail itself of operations/operators
> of t.

join . lift :: m (t m a) -> t m a

morph is more powerful than lift, but it isn't because of the type.

-- 
Dave Menendez <dave at zednenem.com>
<http://www.eyrie.org/~zednenem/>