java - 导致无限循环的异常处理(java)

Question

你好，我是 Java 初学者。我想创建一个方法，该方法采用 1 到 9 之间的整数。我正在使用异常处理程序，以便它可以处理错误或不匹配的输入，但它似乎只执行语句“choice = input.nextInt()”一次，我的循环因此变得无限。

代码如下:

import java.util.*;

public class Player{
    private int[] over;
    private int choice;
    private int coordinates[];
    private static Scanner input = new Scanner(System.in);


    public Player(){
        over = new int[5];
        for(int i = 0; i < 5; i++){
            over[i] = 1;
        }
        coordinates = new int[2];
        coordinates[0] = coordinates[1] = -1;
    }


    public void getChoice(){
            int choice = -1;
            boolean inputIsOk;
            do{
                System.out.print("Enter Your Choice: ");
                inputIsOk = true;
                try{
                    choice = input.nextInt();
                }
                catch(InputMismatchException e){
                    System.out.println("Invalid choice");
                    inputIsOk = false;
                }
                if(choice < 1 || choice > 9){
                    System.out.println("Enter Choice In Range(1-9)");
                    inputIsOk = false;
                }
            }while(!inputIsOk);
            System.out.println("You Entered "+choice);
    }
} 

这是测试类:

public class TestPlayer{
    public static void main(String args[]){
        Player p1 = new Player();
        p1.getChoice();
    }
}

这是输出:第一种情况只输入整数选择时

harsh@harsh-Inspiron-3558:~/java/oxgame$ javac TestPlayer.java 
harsh@harsh-Inspiron-3558:~/java/oxgame$ java TestPlayer 
Enter Your Choice: 10
Enter Choice In Range(1-9)
Enter Your Choice: -1
Enter Choice In Range(1-9)
Enter Your Choice: 55
Enter Choice In Range(1-9)
Enter Your Choice: 5
You Entered 5

第二次当我输入错误的输入时:

Enter Your Choice: 10
Enter Choice In Range(1-9)
Enter Your Choice: 55
Enter Choice In Range(1-9)
Enter Your Choice:g
Enter Your Choice: Invalid choice
Enter Choice In Range(1-9)
Enter Your Choice: Invalid choice
Enter Choice In Range(1-9)
Enter Your Choice: Invalid choice
Enter Choice In Range(1-9)
Enter Your Choice: Invalid choice
Enter Choice In Range(1-9)
Enter Your Choice: Invalid choice
Enter Choice In Range(1-9)
Enter Your Choice: Invalid choice
Enter Choice In Range(1-9)
Enter Your Choice: Invalid choice
Enter Choice In Range(1-9)
Enter Your Choice: Invalid choice
Enter Choice In Range(1-9)
and it goes on....

请帮助我，谢谢。

Answer 1

如果你改变 catch 子句如下:

} catch (InputMismatchException e) {
    input.next();
    System.out.println("Invalid choice");
    inputIsOk = false;
}

这会起作用，input.next(); 我不知道为什么，旧代码 - 当你输入 g - 只是执行这个 choice = input.nextInt(); 就好像它仍然保持相同的值一样，它没有等待用户输入，调用 next () 解决了这个问题。