Hi Jeremy,
On Sun, Aug 23, 2009 at 5:08 PM, Jeremy Shaw
What you probably want is:
test2' :: IO () test2' = runM "foo" $ do loop callback liftIO $ print "here"
This equals my test1 version which is fine without forkIO.
return $ loop callback :: (Monad m) => IO (M ())
It is an IO operation which returns a value of type, M (). But, nothing is done with that value, it is just thrown away.
I see. This is what I feared.
If you want to add a forkIO, the forkIO must go before the runM:
OK. I tried all kinds of combinations, but not forkIO in front of 'runM' . The reason being that I only use forkIO because 'loop' never returns so in order to proceed I would have to put it into the background. Basically I want two threads running inside the same M monad.
hope this helps.
Many thanks Jeremy. Your explanations are indeed very helpful. But they also strengthen my gut feeling that running two threads in the same M monad is hard (if not impossible) to do. Cheers, Levi