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 wrote:

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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Just like sequence, there is also a function sequence_ that you might want to use. sequence_ :: Monad m => [m a] -> m () -- for m = IO, sequence_ :: [IO a] -> IO () sequence_ lst = do sequence lst return () -- also, mapM and mapM_ are useful too mapM f = sequence . map f -- retains results mapM_ f = sequence_ . map f -- ignores results -- Some use cases ghci> sequence [print 2, print 3] 2 3 [(),()] -- The results, each print is of type IO (), i.e does IO and produces () ghci> mapM print [2,3] -- same as above ghci> mapM_ print [2,3] -- same as above, but without results (or result of type unit, i.e ()) On 10 March 2015 at 02:09, Joel Williamson wrote:

sequence will get the types to match up, but a more elegant solution would be to get every line into a single string, then print that.

putStrLn $ unlines $ map show newList

I agree that getting the types to line up can be a nuisance, and with such small programs it doesn't bring much of an advantage. Ultimately, if you want side-effects to be reflected in the type system, there will be times you have to do a bit of extra work to satisfy the type checker. Learning to write well-typed code is much easier in this context, where everything has a fairly concrete type and there is lots of documentation, than having to learn it later when you have weird types coming from 3 different libraries and are facing a problem no one else has had. I would recommend getting very comfortable with GHCi and Hoogle.

If you haven't already, add a hoogle prompt to GHCi by pasting something like :def hoogle \str -> return $ ":! hoogle --count=15 \"" ++ str ++ "\"" in your ghci.conf. This will allow you to easily search for functions of a given type. Typing :hoogle [IO a] -> IO [a] returned all the information needed to answer your question.

...

Thanks, Joel.

Putting the type IO [()] in the main declaration and this as the final

On Mon, Mar 9, 2015 at 4:25 PM, Geoffrey Bays wrote: 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 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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-- Regards Sumit Sahrawat