Question

我想在main中使用累加器。

让我们说：

main = do
   putStrLn "Hey there, what's your name and age?"
   name <- getLine
   age <- getLine
   putStrLn ("Hi " ++ name ++ " you are " ++ age ++ " years old!")
   main

所以我想要做的是如果用户没有提供他的名字和年龄，我想使用之前在下一次主要呼叫中输入的年龄。有什么建议我能做到吗？

由于

Answer 1

确保只使用另一个可以使用您选择的参数的函数：

ask :: String -> IO ()
ask lastAge = do
   -- don't know what you want to do with lastAge
   -- so I just use it when the user did not enter anything
   putStrLn "Hey there, what's your name and age?"
   name <- getLine
   age' <- getLine
   let age = if null age' then lastAge else age'
   putStrLn ("Hi " ++ name ++ " you are " ++ age ++ " years old!")
   ask age

main :: IO ()
main = ask ""

请允许我介绍一个小改进：你可能不希望这里有无限循环 - 你可以使用unless轻松摆脱它：

import Control.Monad (unless)

ask :: String -> IO ()
ask lastAge = do
   -- don't know what you want to do with lastAge but here you have it
   putStrLn "Hey there, what's your name and age?"
   name <- getLine
   unless (null name) $ do
     age' <- getLine
     let age = if null age' then lastAge else age'
     putStrLn ("Hi " ++ name ++ " you are " ++ age ++ " years old!")
     ask age

一旦用户没有提供他的名字

，这将停止循环

在主要中使用蓄能器

1 个答案: