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?
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