list - Haskell IO 创建一个字符串列表并显示它

Question

我正在尝试编写一个程序，允许用户通过一次输入一个字符串来构建一个字符串列表，并在每一步后显示该列表。

到目前为止，这是我的代码:

buildList :: [String] -> IO ()
buildList arr = do
    putStr "Enter a line:"
    str <- getLine
    if str == "" then
        return ()
    else do
        let newarr = arr : str
        putStrLn ("List is now: " ++ newarr)
        buildList newarr


listBuilder :: IO ()
listBuilder = do
    buildList []

listBuilder 通过传入空列表来启动列表，我正在尝试使用递归，以便代码一直运行，直到用户输入空字符串。

它不工作，欢迎任何想法

这是一个需要的输入:

Enter a line: hello
List is now ["hello"]
Enter a line: world
List is now ["hello","world"]
Enter a line:

错误:

Couldn't match type `Char' with `[String]'
Expected type: [[String]]
  Actual type: String
In the second argument of `(:)', namely `str'
In the expression: arr : str
In an equation for `newarr': newarr = arr : str

编辑:

由于线索和 show

的使用，这修复了它

buildList :: [String] -> IO ()
buildList arr = do
    putStr "Enter a line:"
    str <- getLine
    if str == "" then
        return ()
    else do
        let newarr = arr++[str]
        putStrLn ("List is now: " ++ show newarr)
        buildList newarr


listBuilder :: IO ()
listBuilder = do
    buildList []

Answer 1

你可以通过

(a) 使用 arr++[str] 而不是 arr:str 将新字符串放在列表的末尾，因为 : 可以只能像singleThing:list,
那样使用 (b) 将 run-round 拆分为一个单独的函数，并且
(c) 使用 return 传递结果，以便您可以在程序的其他地方使用它

buildList arr = do
    putStrLn "Enter a line:"
    str <- getLine
    if str == "" then
        return arr
    else do
        tell (arr++[str])

tell arr = do
        putStrLn ("List is now: " ++ show arr) -- show arr to make it a String
        buildList arr

给予

Enter a line:
Hello
List is now: ["Hello"]
Enter a line:
world
List is now: ["Hello","world"]
Enter a line:

done