錯誤的代碼片段

Question

有人可以請解釋在下面的代碼片段中的錯誤。在此代碼中，我試圖執行BST的序遍歷，同時從用戶以輸入

import System.IO 

loop :: Int -> [Int] -> IO [Int] 
loop 0 ls = return ls 
loop n ls = do newNumber <- readLn 
       loop (n-1) (newNumber:ls) 

data Tree a = EmptyTree | Node a (Tree a) (Tree a) deriving (Show,Read, Eq) 

singleton :: a -> Tree a 
singleton x = Node x EmptyTree EmptyTree 

treeInsert :: (Ord a) => a -> Tree a -> Tree a 
treeInsert x EmptyTree = singleton x 
treeInsert x (Node a left right) 
        | x==a = Node x left right 
        | x<a = Node a (treeInsert x left) right 
        | x>a = Node a left (treeInsert x right) 

inorder :: Tree a -> [a] 
inorder EmptyTree = [] 
inorder (Node a left right) = inorder left ++ [a] ++ inorder right 

main = do 
    putStrLn " Please enter the number" 
    number <- readLn :: IO Int 
    putStrLn $ "The num is:" ++ show number 
    xs <- loop number [] 
    let numtree = foldr treeInsert EmptyTree xs 
    print numtree 
    ys <- inorder numtree 
    print ys 

我得到的錯誤是：

Couldn't match expected type `IO t0' with actual type `[a0]' 
In the return type of a call of `inorder' 
In a stmt of a 'do' expression: ys <- inorder numtree 
In the expression: 
    do { putStrLn " Please enter the number"; 
     number <- readLn :: IO Int; 
     putStrLn $ "The num is:" ++ show number; 
     xs <- loop number []; 
     .... } 

Answer 1

一些問題，我的代碼看到的是：

1. 你調用一個函數loop你沒有定義或進口任何地方
2. 你沒有縮進的的內容
3. 你寫ys <- inorder numtree，這意味着inorder返回IO something，但它不。您應該使用let。
4. 你do - 嵌段與<- assignemnt結束 - 它應該用表達結束。