Thank you! This works: first :: (Monad a) => ReaderT [String] a String Grzegorz W dniu 2015-10-15 o 23:55, Alexey Shmalko pisze:

Hi,

You could generalize first to ReaderT [String] a String, so that you could use it as ReaderT [String] IO String. You don't need to change the implementation - just change the type.

Hope this helps, Alexey

On Fri, Oct 16, 2015 at 12:37 AM, Grzegorz Balcerek wrote:

...

Hi,

I have the following program:

import Control.Monad.Reader

first :: Reader [String] String first = do strings <- ask return $ if (null strings) then "empty" else head strings

printFirst :: ReaderT [String] IO () printFirst = do strings <- ask let theFirstString = runReader first strings liftIO $ putStrLn theFirstString

main = runReaderT printFirst ["first","second"]

It compiles and works. However, in the printFirst function I am explicitly using ask and I am calling runReader. Can I somehow avoid doing that? The following version of the printFirst function does not compile.

printFirst :: ReaderT [String] IO () printFirst = do theFirstString <- first liftIO $ putStrLn theFirstString

Program2.hs:11:21: Couldn't match type `Data.Functor.Identity.Identity' with `IO' Expected type: ReaderT [String] IO String Actual type: Reader [String] String In a stmt of a 'do' block: theFirstString <- first In the expression: do { theFirstString <- first; liftIO $ putStrLn theFirstString }

Can I somehow call first without using ask and runReader ?

Regards Grzegorz Balcerek

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