CODEWITHSUNDEEP
haskellexceptionio
KaramJaber
KaramJaber

Reputation: 893

*** Exception: user error (Prelude.readIO: no parse) in haskell

I'm trying to write a program that loops all the time waiting for an input from the user but for some reason it doesn't loop. My program is :

charAt :: String->Char->Int
main = do 
    x <- readLn
    if x == 1
        then do
           putStrLn "Word: "
           word <- getLine
           putStrLn "Char: "
           ch <- getChar
           putStrLn (show (charAt word ch))
        else print "Nothing"
    main

But i actually get this error:

*** Exception: user error (Prelude.readIO: no parse)

If i remove the last main calling the program will work . does anybody knows why is that happening?

Upvotes: 1

Views: 3446

Answers (1)

Zeta
Zeta

Reputation: 105876

When you use getChar, it will take only a single character from your stream. However, if you've entered AEnter, the newline character '\n' is still in your stdin. A '\n' cannot get parsed into an Int, therefore you end up with that error.

You can remove that newline if you call getLine afterwards:

       ch <- getChar
       putStrLn (show (charAt word ch))
       _ <- getLine

Or you write your own helpers:

promptInt :: IO Int
promptInt = do
  putStr "Int: "
  line <- getLine
  case readMaybe line of
    Just x -> return x
    _      -> putStrLn "Try again." >> promptInt

promptChar :: IO Char
promptChar = do
  putStr "Char: "
  line <- getLine
  case line of
    [x,_] -> return x
    _     -> putStrLn "Try again." >> promptChar

Upvotes: 4

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