Move MonadIO to base

[wren ng thornton]

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))


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.

-- 
Live well,
~wren