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
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