Is there a way to unwrap a type from an IO monad?

haskell, io, io-monad, monads

Solution

To expand on my comments, here's how you can do it:

getJson :: IO B.ByteString
getJson = B.readFile jsonFile -- as before

readJFile :: B.ByteString -> Maybe Response -- look, no IO
readJFile b = case eitherDecode b of
                Left err -> Nothing
                Right ps -> Just ps

In the end, you combine everything in one IO action again:

getAndProcess :: IO (Maybe Response)
getAndProcess = do
  b <- getJson
  return (readJFile b)

Problem

I have this very simple function ``` import qualified Data.ByteString.Lazy as B getJson :: IO B.ByteString getJson = B.readFile jsonFile readJFile :: IO (Maybe Response) readJFile = parsing >>= (\d -> case d of Left err -> return Nothing Right ps -> return (Just ps)) where parsing = fmap eitherDecode getJson :: IO (Either String Response) ``` where `jsonFile` is a path to a file on my harddrive (pardon the lack of do-notation, but I found this more clear to work with) my question is; is there a way for me to ditch the `IO` part so I can work with the bytestring alone? I know that you can pattern match on certain monads like `Either` and `Maybe` to get their values out, but can you do something similar with `IO`? Or voiced differently: is there a way for me to make `readJFile` return `Maybe Response` without the IO?

Original source