How to run official example of Haskell WebSockets library - haskell

Question

According to this topic: https://stackoverflow.com/questions/14494922/examples-of-websocket-usage-in-haskell I have my first question. Why official example of WebSockets library doesn't run on my machine?
import Data.Char (isPunctuation, isSpace)
import Data.Monoid (mappend)
import Data.Text (Text)
import Control.Exception (fromException)
import Control.Monad (forM_, forever)
import Control.Concurrent (MVar, newMVar, modifyMVar_, readMVar)
import Control.Monad.IO.Class (liftIO)
import qualified Data.Text as T
import qualified Data.Text.IO as T
import Network.WebSockets
meow :: TextProtocol p => WebSockets p ()
meow = forever $ do
msg <- receiveData
sendTextData $ msg `T.append` ", meow."
main :: IO ()
main = runServer "0.0.0.0" 8000 meow
I get:
ciembor#freedom ~> ghc hchat.hs; and ./hchat
[1 of 1] Compiling Main ( hchat.hs, hchat.o )
hchat.hs:15:35:
Couldn't match expected type `Text' with actual type `[Char]'
In the second argument of `T.append', namely `", meow."'
In the second argument of `($)', namely `msg `T.append` ", meow."'
In a stmt of a 'do' block: sendTextData $ msg `T.append` ", meow."
hchat.hs:18:33:
Couldn't match expected type `Request -> WebSockets p0 ()'
with actual type `WebSockets p1 ()'
In the third argument of `runServer', namely `meow'
In the expression: runServer "0.0.0.0" 8000 meow
In an equation for `main': main = runServer "0.0.0.0" 8000 meow

Answer 1

The first error is because you didn't enable the OverloadedStrings language extension. Without that, a "xyz" is a String, but Data.Text.append takes two Texts as arguments. Add
{-# LANGUAGE OverloadedStrings #-}
to the top of the module to fix that one.
The second error is because the third argument to runServer must have type
Request -> WebSockets p0 ()
but you gave it a WebSockets p1 (). If you want to pass an action, you have to lift it to a function,
main = runServer "0.0.0.0" 8000 $ const meow
would compile (whether it would work [do what you want] is a question I cannot answer, that would just ignore all requests and always do the same thing).