On Sun, 18 Apr 2010, wren ng thornton wrote:
> lift ma = morph (\k -> join (fmap (k . return) ma))
Monad laws simplify that to
lift ma = morph (\k -> ma >>= k . return)
> The type of morph requires us to Church-encode things needlessly; what
> we mean to say is: morph (fmap return ma).
Hmm. If I understand this (and your other emails) correctly, you’re
saying my interface
class Monad m => MonadMorphIO m where
morphIO :: (forall b. (m a -> IO b) -> IO b) -> m a
should be equivalent to a simpler interface
class Monad m => MonadJoinIO m where
joinIO :: IO (m a) -> m a
via some isomorphism like
morphIO f = joinIO (f return)
joinIO m = morphIO (\w -> m >>= w)
I would be very happy to get the simpler interface to work, because it’s
Haskell 98. However, if I write
joinIO m = morphIO (\w -> m >>= w)
morphIO' f = joinIO (f return)
and define catch using morphIO' instead of morphIO:
m `catch` h = morphIO $ \w -> w m `Control.Exception.catch` \e -> w (h e)
m `catch'` h = morphIO' $ \w -> w m `Control.Exception.catch` \e -> w (h e)
then catch' fails to actually catch anything:
*Main> throwIO NonTermination `catch` \NonTermination -> return "moo"
"moo"
*Main> throwIO NonTermination `catch'` \NonTermination -> return "moo"
*** Exception: <<loop>>
Am I doing something wrong?
Anders
