如何通过IO操作运行列表

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));
}

但是我需要在整个列表中运行singleWord而不是singleWord (head (lines content)) 。

问题是，对于map或普通列表理解，我总是会遇到大量不同的错误，这些错误与lines content （应该是字符串或IO字符串数组）有关，显然总是我不想要的类型（我是我一直试图弄乱类型声明，但它仍然是错误的类型，或者是正确的类型，但在额外的数组层或其他方面）。

我的最后一次尝试是通过递归来遍历数组，这个额外的代码：

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

无论如何我无法正确地进行类型转换。

它应该更换singleWord (head (lines content))在main ，如果我尝试任何与typeclassing，像

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 ()

或其他一些不适合的东西。

Answer 1

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 2

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) }

或者更好，

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编写它，

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

sequence_ :: Monad m => [ma] -> m ()将一个动作列表转换为一系列动作，丢弃它们的结果并在最后生成() ： sequence_ = foldr (>>) (return ()) 。 而sequence_ (map f xs) === mapM_ f xs ，所以它们最终都会mapM_ f xs 。

Answer 3

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

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