如何在Haskell中打印列表的单个元素？

我是 Haskell 的初学者，我想知道是否可以打印给定列表的单个元素。我尝试解决这个问题，但失败了。这是代码：

main :: IO()
main = do
    let list = [1,2,3]
    let size = length list
    let temp = print_elem list size
    print temp


print_elem :: [Int] -> Int -> Int
print_elem xs x = do
    let size = length xs
    let idx = size - x
    let element = xs !! idx
    putStr (show element)
    putStr(" ")
    let dummy = print_elem (xs (x-1))
    return " "

我想打印这样的东西

1 2 3

如果我只是使用putStr (show list)它会显示[1,2,3]，我不想要那样。

但是当我运行这段代码时会出现多个错误

printelem.hs:14:5: error:
    * Couldn't match expected type `Int' with actual type `IO b0'
    * In a stmt of a 'do' block: putStr (show element)
      In the expression:
        do let size = length xs
           let idx = size - x
           let element = xs !! idx
           putStr (show element)
           ....
      In an equation for `print_elem':
          print_elem xs x
            = do let size = ...
                 let idx = ...
                 let element = ...
                 ....
   |
14 |     putStr (show element)
   |     ^^^^^^^^^^^^^^^^^^^^^

printelem.hs:16:29: error:
    * Couldn't match expected type `Int -> [Int]'
                  with actual type `[Int]'
    * The function `xs' is applied to one argument,
      but its type `[Int]' has none
      In the first argument of `print_elem', namely `(xs (x - 1))'
      In the expression: print_elem (xs (x - 1))
   |
16 |     let dummy = print_elem (xs (x-1))
   |                             ^^^^^^^^

你如何解决这个问题？

回答

> map show [1,2,3]
["1","2","3"]

> import Data.List
> intercalate " " (map show [1,2,3])
"1 2 3"

import Data.List

main :: IO()
main = do
    let list = [1,2,3]
    putStrLn (intercalate " " (map show list))