如何在 Haskell 中打印列表的各个元素?

How to print individual elements of a list in Haskell?

我是 Haskell 的初学者,我想知道是否可以打印给定列表的各个元素。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

如果我简单地使用 putStr (show list) 它将显示 [1,2,3] 而我不想要那个。

但是当我运行这段代码出现多个错误

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?

你想得太急迫了。首先,您需要一个字符串列表,而不是一个整数列表。那是 map:

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

接下来,您想将它们连接成一个单独的 space 字符串。那是 Data.List.intercalate:

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

然后您可以将其传递给 putStrLnprint 将为您提供已有字符串的字符串表示形式):

import Data.List

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

一种方法是

> mapM_ putStr (intersperse " " (map show [1,2,3])) >> putStrLn ""
1 2 3
it :: ()

这会逐一打印元素,通过 show 转换为字符串,中间穿插空格。

> intersperse " " (map show [1,2,3])
["1"," ","2"," ","3"]
it :: [[Char]]

mapM_ 在每个元素上映射一个 IO 动作构造函数 putStr :: String -> IO (),并将它们作为一个组合的动作序列执行。

最后这段代码打印换行符。