Haskell 中有用的 ReadLn

Question

Haskell 中是否有像 Pascal 中的 ReadLn 这样的内置函数？

我想要这样的东西：

λ> pascalReadLn :: IO (Int, Int, Int, Int)
1 2
3
4
(1,2,3,4)
λ> pascalReadLn :: IO (Int, Int, Int, Int)
1 2 3 4
(1,2,3,4)
λ> pascalReadLn :: IO (Int, Int, Int, Int)
1
2
3
4
(1,2,3,4)
...
etc.

Answer 1

您可以使用ReadArgs 一起破解它

import ReadArgs

pascalReadLn :: ArgumentTuple a => IO a
pascalReadLn = pascalReadLn' ""
  where pascalReadLn' lines = do
          line <- getLine
          let lines' = lines ++ line
          -- see if we've read enough to successfully parse the desired tuple
          case parseArgsFrom (words lines') of
            Nothing -> pascalReadLn' (lines' ++ "\n")
            Just a  -> return a

它可以根据需要进行有效输入

λ pascalReadLn :: IO (Int, Int, Int, Int)
1 2
3
4
(1,2,3,4)
λ pascalReadLn :: IO (Int, Int, Int, Int)
1 2 3 4
(1,2,3,4)
λ pascalReadLn :: IO (Int, Int, Int, Int)
1
2
3
4
(1,2,3,4)

然而，它并不完美，因为它无法区分不完整的解析和不可能的解析：

λ pascalReadLn :: IO (Int, Int, Int, Int)
foo bar
1
2
3
4
... will go forever

自定义实现（与ArgumentTuple 相同）可以通过区分两种失败情况来解决此问题，例如：

 data ParseResult a = Success a | Incomplete (String -> ParseResult a) | Impossible

 class LineTuple a where
   parseLine :: String -> ParseResult a