I have been working through the exercises in a Write Yourself a Scheme in 48 Hours/Parsing and hacked together something to get parseString to comprehend escaped characters. Also had some inspiration from Real World Haskell chapter 16 minus the applicative parts.parseString :: Parser LispValparseString = do char ''                 x <- many $ chars                 char ''                 return $ String x    where chars = escaped <|> noneOf \          escaped = choice $ map tryEscaped escapedChars          tryEscaped c = try $ char '\\' >> char (fst c) >> return (snd c)          escapedChars = zip bnfrt\\\/ \b\n\f\r\t\\\/This works but I am not fond of my 'tryEscaped' definition. What are the alternatives?
Parsing strings with escaped characters using Parsec
haskell;parsec
Some suggestions:Replace fst and snd with a pattern match or explicit function arguments.Extract the common char '\\' parser. You can then avoid the try.Strings with lots of escaped characters are hard to read and it's hard to visuallyverify that the escape codes are correctly matched with their replacements. Considerspelling them out and aligning them to make this easy to see. Here's what I came up with:escaped = char '\\' >> choice (zipWith escapedChar codes replacements)escapedChar code replacement = char code >> return replacementcodes        = ['b',  'n',  'f',  'r',  't',  '\\', '\', '/']replacements = ['\b', '\n', '\f', '\r', '\t', '\\', '\', '/']