Am Samstag 23 Januar 2010 21:31:47 schrieb Adrian Adshead:
Hi Alex,
That does help, but raises a further question.
If (>>=) :: m a -> (a -> m b) -> m b
Then whenever the IO monad is entered it can not be escaped, so in the example you gave:-
f :: IO a -> Int f _ = 10
while f (putStrLn "foo") is a pure function evaluating to 10, neither the IO Action, nor the a (from 'IO a') can be used without using the (>>=) or (>>). Therefore since the parameter can not be used in the function f without entering the IO monad, your function would be equivalent to :-
f :: Int f = 10
Rather to const 10 except with a restricted type.
I have two questions to try and clarify my understanding.
1) Is it possible to have a function with an IO Action as an input parameter (and not just ignore the parameter) and have a non-IO return value? (ie is f :: IO a -> b possible without ignoring the IO a)
Excluding unsafePerformIO, only in a very weak sense. ioseq :: IO a -> b -> b ioseq ioa b = ioa `seq` b doesn't quite ignore the parameter, but apart from that, I only see rustyBallpointTracheotomy (Kydos to Brent Yorgey for that :).
2) Is it possible to have a function which evaluates any IO Action within the body and have a non-IO return value? (ie is f :: a -> b possible while evaluating any IO Action within the body.)
Assuming by evaluating you mean executing and not (return 5 >> return 6) ~> IO (\s -> (s,6)) , no, that's not possible.
My understanding of Haskell was that both of these situations would not be possible, and the return value would always be of the form IO b. (For these questions I am excluding use of unsafePerformIO)
Many thanks for your time.
Adrian.