If I modify the code to the suggested form: let f = runWorker (worker path) in do f loop then I get the following compile error Couldn't match type `IO' with `InputT IO' Expected type: InputT IO () Actual type: IO () In a stmt of a 'do' block: f In the expression: do { f; loop } I have also tried let f = runWorker (worker path) in do return f loop This compiles, but won't run forkIO. I can't figure out how to get forkIO to run and make the type system happy. I am going to past the code again, because the line breaks got messed up the first time. I have removed the `seq` call though as it simplifies the code a bit. import Control.Concurrent (forkIO) import Control.Exception import qualified Data.ByteString.Lazy as L import System.Console.Haskeline hiding (handle) -- Provided by the 'zlib' package on http://hackage.haskell.org/ import Codec.Compression.GZip (compress) worker :: FilePath -> IO () worker path = L.readFile path >>= L.writeFile (path ++ ".gz") . compress runWorker :: FilePath -> IO() runWorker path = handle (print :: SomeException -> IO ()) $ do forkIO (worker path) return () loop :: InputT IO () loop = do maybeLine <- getInputLine "Enter a file to compress> " case maybeLine of Nothing -> return () -- user entered EOF Just "" -> return () -- treat no name as "want to quit" Just path -> do return (runWorker path) loop main = runInputT defaultSettings loop

Thanks Peter, That suggestion works. I will have to continue learning more about Monads. --Jeff -----Original Message----- From: Beginners [mailto:beginners-bounces@haskell.org] On Behalf Of Peter Jones Sent: Thursday, August 28, 2014 7:11 AM To: beginners@haskell.org Subject: Re: [Haskell-beginners] Haskeline and forkIO "Jeff C. Britton" writes:

loop :: InputT IO () loop = do maybeLine <- getInputLine "Enter a file to compress> " case maybeLine of Nothing -> return () -- user entered EOF Just "" -> return () -- treat no name as "want to quit" Just path -> do return (runWorker path) loop

The other issue you're having is because `runWorker path` is an `IO ()` value but at the point where you use it in the code the type system wants an `InputT IO ()`. To try to satisfy the type system you used `return` to build a `InputT IO (IO ())` value, but that doesn't actually work (as you've noticed). Since `InputT` is a transformer you have an extra layer to work through and so need to *lift* your `IO ()` value into the `InputT IO` layer. Try this: -- Add this import import Control.Monad.IO.Class loop :: InputT IO () loop = do maybeLine <- getInputLine "Enter a file to compress> " case maybeLine of Nothing -> return () -- user entered EOF Just "" -> return () -- treat no name as "want to quit" Just path -> do liftIO (runWorker path) loop You can think of `liftIO` as having this signature (in this context): liftIO :: IO () -> InputT IO () -- Peter Jones, Founder, Devalot.com Defending the honor of good code _______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

As a Haskell beginner, I would also like to recommend HLint as a very useful tool to catch these kinds of things. On Sep 3, 2014 12:25 PM, "Peter Jones" wrote:

"Jeff C. Britton" writes:

...

That suggestion works. I will have to continue learning more about Monads.

I should have also mentioned that if you enable GHC warnings it should tell you that having the `return` before `loop` is discarding the value given to it.

...

-----Original Message----- From: Beginners [mailto:beginners-bounces@haskell.org] Sent: Thursday, August 28, 2014 7:11 AM To: beginners@haskell.org Subject: Re: [Haskell-beginners] Haskeline and forkIO

"Jeff C. Britton" writes:

...

loop :: InputT IO () loop = do maybeLine <- getInputLine "Enter a file to compress> " case maybeLine of Nothing -> return () -- user entered EOF Just "" -> return () -- treat no name as "want to quit" Just path -> do return (runWorker path) loop

The other issue you're having is because `runWorker path` is an `IO ()` value but at the point where you use it in the code the type system wants an `InputT IO ()`. To try to satisfy the type system you used `return` to build a `InputT IO (IO ())` value, but that doesn't actually work (as you've noticed). Since `InputT` is a transformer you have an extra layer to work through and so need to *lift* your `IO ()` value into the `InputT IO` layer. Try this:

-- Add this import import Control.Monad.IO.Class

loop :: InputT IO () loop = do maybeLine <- getInputLine "Enter a file to compress> " case maybeLine of Nothing -> return () -- user entered EOF Just "" -> return () -- treat no name as "want to quit" Just path -> do liftIO (runWorker path) loop

You can think of `liftIO` as having this signature (in this context):

liftIO :: IO () -> InputT IO ()

-- Peter Jones, Founder, Devalot.com Defending the honor of good code

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