我如何运行通过一个 IO 操作的列表

Question

所以我正在编写一个程序来检查 .txt 文件的每一行是否是回文，

import System.IO

main :: IO()
main = do {
content <- readFile "palindrom.txt";
print content;
print (lines content);
singleWord (head (lines content));
return ();
}

palindrom :: [Char] -> Bool
palindrom a = a == reverse a

singleWord :: [Char] -> IO()
singleWord a = do { 
print (length a);
print (show (palindrom a));
}

但我需要运行 singleWord 遍历整个列表，而不是 singleWord (head (lines content))。

问题是，使用 map 或正常的列表理解，我总是会遇到大量与 lines content（应该是字符串或 IO 字符串数组）有关的各种错误，显然总是是我不想要的类型（我试过永远弄乱类型声明，但它一直是错误的类型，或者是正确的类型，但在额外的数组层或其他什么）。

我的最后一次尝试是使用递归遍历数组，使用这个额外的代码：

walkthrough [] = []
walkthrough x = do { singleWord head x; walkthrough (tail x) }

无论如何我都无法正确输入。

它应该替换 main 中的 singleWord (head (lines content))，如果我尝试使用类型分类，比如

walkthrough :: [[Char]] -> [[Char]]
walkthrough [] = ["Hi"]
walkthrough x = do { singleWord head x; walkthrough (tail x) }

我明白了

Couldn't match type `IO' with `[]'
      Expected type: [()]
        Actual type: IO ()

或其他一些不能组合在一起的东西。

Answer 1

使用mapM_ singleWord (lines content)。为了简单起见，将 mapM_ 视为。

mapM_ :: (a -> IO ()) -> [a] -> IO ()

Answer 2

您正在寻找一个名为 mapM_ 的函数。

main :: IO ()
main = do {
  content <- readFile "palindrom.txt";
  mapM_ singleWord (lines content);
};


palindrome :: [Char] -> Bool
palindrome a = (a == reverse a)

singleWord :: [Char] -> IO()
singleWord a = do {
  let {
    adverb = (if palindrome a then " " else " not ");
  };
  putStrLn  (a ++ " is" ++ adverb ++ "a palindrome.");
};

Answer 3

那应该是

walkthrough [] = return ()                    -- this is the final action
walkthrough x  = do { singleWord (head x)     -- here you missed the parens 
                    ; walkthrough (tail x) }

或更好，

walkthrough []     = return () 
walkthrough (x:xs) = do { singleWord x        -- can't make that mistake now!
                        ; walkthrough xs}

并在主 do 块中将其称为 walkthrough (lines content)。

正如其他人所指出的，walkthrough 与 mapM_ singleWord 相同。

你也可以用列表理解来写，

walkthrough xs = sequence_ [ singleWord x | x <- xs]

sequence_ :: Monad m => [m a] -> m () 将动作列表变成一系列动作，丢弃它们的结果并最终生成 ()：sequence_ = foldr (>>) (return ())。而sequence_ (map f xs) === mapM_ f xs，所以最后一切都结束了。

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