Haskell - 为什么它不起作用？ （IO行动列表）

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

我在某个网站上看到了这个例子（下面），我想知道它是如何工作的，所以我在其中放了一些IO动作。但是，ghci给了我一个类型错误。有什么问题？

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

Answer 1

主要问题是map和++适用于[a]，而非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