[Haskell-cafe] Re: Move MonadIO to base
wren ng thornton
wren at freegeek.org
Sun Apr 18 17:17:22 EDT 2010
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.
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