java - 异常处理导致无限循环(java)

您好，我是Java初学者。我想制作一个采用1到9之间的整数的方法。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，但似乎只执行语句“choice”。 = input.nextInt()”一次，因此我的循环进入了无限。

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();
    }
}

输出是:
第一种情况当仅输入积分选择时

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

其次，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。

最佳答案

这会工作

try{
       choice = Integer.parseInt(input.next());
   }
catch(NumberFormatException e){
    System.out.println("Invalid choice");
    inputIsOk = false;
}

原因是:
说扫描程序从流中读取一个对象，说typeCache。直到不会获得整数值，缓冲区才会被刷新，并且typeCache将保存String，直到使用next()(或任何等效方法)读取它为止。
Scanner类的代码:

    public int nextInt() {
    return nextInt(defaultRadix);
}

public int nextInt(int radix) {
    // Check cached result
    if ((typeCache != null) && (typeCache instanceof Integer)
        && this.radix == radix) {
        int val = ((Integer)typeCache).intValue();
        useTypeCache();
        return val;
    }.......

要么
只需在您的catch块中添加input.next();，它将自动清除typeCache。