Haskell - 为什么它不起作用？ （列表中的IO动作）

Question

combinationIO :: Int -> [a] -> IO [[a]] 
combinationIO 0 _ = return [[]] 
combinationIO _ [] = return [] 
combinationIO n (x:xs) = do res <- (map (x:) (combinationIO (n-1) xs)) ++ (combinationIO n xs) 
          putStrLn $ (show n) ++ show " : (" ++ show (x,xs) ++ show ") = " ++ show res 
          return res 

我在一些网站上看到了这个例子（下面），我想知道它是如何工作的，所以我在里面放了一些IO动作。但是，ghci给我一个类型错误。问题是什么？

combination2 :: Int -> [a] -> [[a]] 
combination2 0 _ = [[]] 
combination2 _ [] = [] 
combination2 n (x:xs) = (map (x:) (combination2 (n-1) xs)) ++ (combination2 n xs) 

Answer 1

的主要问题是，map和++工作的[a]，不IO [a]。我想你想要的是这样的：

combinationIO :: Show a => Int -> [a] -> IO [[a]] 
combinationIO 0 _ = return [[]] 
combinationIO _ [] = return [] 
combinationIO n (x:xs) = do 
    res1 <- combinationIO (n-1) xs 
    res2 <- combinationIO n xs 
    let res = (map (x:) res1) ++ res2 
    putStrLn $ (show n) ++ " : (" ++ (show (x,xs)) ++ ") = " ++ (show res) 
    return res