Question

我一直在写我的第一个Java脚本，遇到了一个问题。我不明白为什么在运行此程序时，该程序会在第一次运行循环时等待用户输入某些内容，然后处理输入并做出相应的响应，但是在第二次运行时，它将卡在无限循环，无需等待用户的输入。作为记录，我确实希望这是一个无限循环，只是如果用户输入“ 3”，则程序应结束。如果在每种情况下我都写“ keep_going = false;”该程序可以运行，但显然不会循环。感谢所有帮助，谢谢！

import java.io.*; 

class Choice
{ 
public static void main (String[] args) 
{  
    String input = ""; 
    Boolean keep_going = true;
    while (keep_going)
    {
        input = "";
        System.out.println("Welcome to my program! Would you like to:");
        System.out.println("1. Say hi.");
        System.out.println("2. Find out my favourite colour.");
        System.out.println("3. End the program.");
        System.out.println(">");
        System.out.print( "> " );
        InputStreamReader isr = new InputStreamReader( System.in );
        BufferedReader buffer = new BufferedReader( isr );
        try
        { 
            input = buffer.readLine();
            buffer.close() ; 
        } 
        catch (IOException e ) 
        { 
            System.out.println(e);  
        } 
        switch (input)
        {
            case "1": System.out.println("Hi!"); break;
            case "2": System.out.println("My favourite colour is blue!"); break;
            case "3": return;
            default : System.out.println(input + " is not a valid option. Please try again.");
        }
    }


}

}

Answer 1

无限循环归因于buffer.close() ;行

进行此调整

删除buffer.close() ;

//buffer.close() ;
case "3": keep_going=false;break;

在Java中接受用户输入时修复无限循环

1 个答案: