Source: Hutton, Graham. "Programming in Haskell" (p. 180)

6. Using getCh, define an action readLine :: IO String that behaves in the same way as getLine, except that it also permits the delete key to be used to remove characters.

Hint: the delete character is ’\ DEL’, and the control character for moving the cursor back one space is ’\b’.

I solved this exercise using one '\b' character, but found online that a solver used two. Why the solver of this problem uses "\b \b" instead of "\b" ? Seems like a typo but I am unsure. I have found it works with three '\b' characters.

How does this character work ?

import System.IO

getCh :: IO Char
getCh = do
  hSetEcho stdin False
  x <- getChar
  hSetEcho stdin True
  return x

readLine :: IO String
readLine = readLine' ""

readLine' :: String -> IO String
readLine' xs = do
  x <- getCh
  case x of
    '\n' -> do
      putChar '\n'
      return xs
    '\DEL' ->
      if null xs
        then readLine' ""
        else do
          putStr "\b \b"
          readLine' (init xs)
    _ -> do
      putChar x
      readLine' (xs    [x])

CodePudding user response：

\b moves the cursor one character back, but it doesn't erase it (at least, not on most terminals). For example, the string abcde\b will be displayed as abcde, and the string abcde\bf as abcdf. That's why the sequence \b \b explicitly overwrites the last character with a space, and then moves the cursor back again.

CodePudding user response：

If you use just "\b", the cursor goes to the left but doesn't delete the character seen there until you overwrite it with a new key input. For example, FOO⟵ will end you up with fo