
There is a difference between IO [()] and IO () and [IO ()]
A type of [IO ()] is a list of actions, none of which have actually been
executed.
A type of IO [()] is a single action that has executed and returned a bunch
of nils.
sequence is one way to combine a list of actions into a single action that
returns a list of their results, but it might be better to try and separate
the pure and impure part of that line of code:
mapM putStrLn $ ([show s | s <- [1,2,3]] :: [String]) :: IO [()]
The type annotations are for explanation only. Then use mapM_ if you do
not want to save these nils for some reason (there are performance
implications).
On Mon, Mar 9, 2015 at 4:25 PM, Geoffrey Bays
Thanks, Joel.
Putting the type IO [()] in the main declaration and this as the final line of the main function does do the trick:
sequence [putStrLn $ show s | s <- newList]
But this is the kind of thing that makes Haskell types difficult for beginners to work with...
Geoffrey
On Mon, Mar 9, 2015 at 4:15 PM, Joel Williamson < joel.s.williamson@gmail.com> wrote:
main must have type IO a. Hoogle tells me that to convert [IO a] -> IO [a], you should use the function sequence. Try applying that to your final line.
On Mon, 9 Mar 2015 16:07 Geoffrey Bays
wrote: My main function looks like this:
main :: [IO()] main = do let stud1 = Student {name = "Geoff", average = -99.0, grades = [66,77,88]} let stud2 = Student {name = "Doug", average = -99.0, grades = [77,88,99]} let stud3 = Student {name = "Ron", average = -99.0, grades = [55,66,77]} let studList = [stud1,stud2] let newList = calcAvg studList [putStrLn $ show s | s <- newList] --putStrLn $ show (newList !! 0) --putStrLn $ show (newList !! 1)
With this final line, putStrLn $ show (newList !! 0), the type IO () in the function declaration compiles fine. But with [putStrLn $ show s | s <- newList] as the final line, [IO ()] in the function declaration will not compile, I get this error:
Couldn't match expected type `IO t0' with actual type `[IO ()]'
What does the declared type need to be for a final line of: [putStrLn $ show s | s <- newList] ???
Thanks,
Geoffrey
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