You cannot sequence two operations from different monads... p has type: m (IO ()) id has type, IO () (in this case because this is what p returns)... You can do: p :: (Monad m) => m (IO ()) p = q >>= (\a -> return a) Or p :: (Monad m) => m (IO ()) p = run q >>= id -- provided an overloaded definition of run is provided for 'm' Keean. Ashley Yakeley wrote:
I suspect someone's come across this before, so maybe there's an explanation for it.
This does not compile:
module Bug where { p :: IO (); p = q >>= id;
q :: (Monad m) => m (IO ()); q = return p; }
Bug.hs:3: Mismatched contexts When matching the contexts of the signatures for p :: IO () q :: forall m. (Monad m) => m (IO ()) The signature contexts in a mutually recursive group should all be identical When generalising the type(s) for p, q
The code looks correct to me. Why must the signature contexts be identical in this case?