Haskell Io 字符串转换

Question

obrob fp =  do
    a <- [(!!) readData fp 0]
    b <- [(!!) readData fp 2]
   return a --(read a :: Int ,read b::[[Int]] )

我从我得到的文件中读取数据

["6",
 "",
 "[[1,2,3,4,5,6],[7,8,9,10,11,12],[13,14,15,16,17,18],[19,20,21,22,23,24],[25,26,27,28,29,30],[31,32,33,34,35,36]]"
]

readData 返回这个。是io字符串列表

但是现在我想从这个列表中取出第一个和第三个元素并返回

(6,
 [[1,2,3,4,5,6],[7,8,9,10,11,12],[13,14,15,16,17,18],[19,20,21,22,23,24],[25,26,27,28,29,30],[31,32,33,34,35,36]]
)

没有 Io 类型。我不想一直使用 monad。

Answer 1

事实上，你无法真正“摆脱” IO。但这不是您刚接触 Haskell 时的问题。看看main的类型：

main :: IO ()

您的整个程序（或任何程序）是一个正在评估的大型 IO 操作。我们尝试做的是在此过程中进行大量纯计算。

这个怎么样（如果这只会使问题更加混乱，我深表歉意）：

-- Simulate your file read operation
readData :: IO [String]
readData = return ["6","","[[1,2,3,4,5,6],[7,8,9,10,11,12],
    [13,14,15,16,17,18],[19,20,21,22,23,24],[25,26,27,28,29,30],
    [31,32,33,34,35,36]]"]


-- pure function - not IO
someOtherFunction (x, ys) = (x > 0, length ys)


obrob :: IO (Bool, Int)
obrob = do
   -- Pattern match out the 1st and 3rd elements
   (a:_:b:_) <- readData

   -- t is the tuple you're trying to get to
   let t = ((read a) :: Int, (read b) :: [[Int]])
   print t 

   -- And inside this do block, that t is not in IO.
   -- Lets pass it to a pure function:
   let u = someOtherFunction t

   -- Later we have to evaluate to something in IO.
   -- It cannot be escaped.
   return u

Answer 2

读心术的微弱尝试：

obrob fp :: Integral i, Read i => (i, [[i]])
obrob fp = let xs = readData fp
           in (read $ readData fp !! 0, read $ readData fp !! 2)

我假设您的 do 语句使用的是单子版本的 list...我不太确定。您需要提供有关类型的更多详细信息，例如 readData、fp 等。