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