[Haskell-cafe] Re: Move MonadIO to base

[wren ng thornton]

This bounced because I have different emails registered for cafe@ and 
libraries@, so forwarding it along to the cafe.


wren ng thornton wrote:
> wren ng thornton 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))
> 
> Or rather,
> 
>     lift ma = morph (\k -> join (fmap (k . return) ma))
> 
> That's what I get for typing without checking. The type of morph 
> requires us to Church-encode things needlessly; what we mean to say is: 
> morph (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