Haskell Attoparsec无限回路

Question

代码基于Haskell Attoparsec，当我使用parseOnly pString "v"时，它给出了正确的答案：Right (DontNeedTrim, "v")。

当我使用parseOnly (many' pString) "v"指令时，它似乎会掉进infinite loop中，最后由于溢出堆栈而失败。

data Signal = NeedTrim
              | DontNeedTrim
              deriving (Show)

pString :: Parser (Signal, [Char])
pString = ((char '\"' *> many' pChar' <* char '\"') >>= \s -> return (NeedTrim, s))
          <|> (many' pChar >>= \s -> return (DontNeedTrim, s))

pChar :: Parser Char
pChar = char '\\' *> (pEscape <|> spaces *> endOfLine *> pChar)
        <|> satisfy (`C.notElem` "\"\\\n#;")

pChar' :: Parser Char
pChar' = char '\\' *> pEscape
         <|> satisfy (`C.notElem` "\\\"")

pEscape :: Parser Char
pEscape = choice (zipWith decode "bnt\\\"" "\b\n\t\\\"")
  where decode c r = r <$ char c

Answer 1

pString中的第二个选项接受空字符串：many' pChar >>= \s -> return (...)。因此，many' pString一直在无限地使用空字符串。