异常处理无限循环

Question

我的问题很简短。我不明白为什么我的程序在捕获错误时会无限循环。我做了一个新的try-
catch语句，但是它循环了，甚至复制，粘贴并修改了以前有效的程序中的适当变量。下面是语句本身，下面是整个程序。谢谢您的帮助！

try {
    input = keyboard.nextInt();
}
catch(Exception e) {
    System.out.println("Error: invalid input");
again = true;

}
if (input >0 && input <=10)
    again = false;

}

程序：

    public class Blanco {

    public static int input;

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here
        nameInput();



    }

    /**
     *
     * @param name
     */
    public static void nameInput() {

        System.out.println("What is the name of the cartoon character : ");
        Scanner keyboard = new Scanner(System.in);
        CartoonStar star = new CartoonStar();
        String name = keyboard.next();
        star.setName(name);
        typeInput(keyboard, star);

    }

    public static void typeInput(Scanner keyboard, CartoonStar star) {

boolean again = true;
while(again){
        System.out.println("What is the cartoon character type: 1 = FOX,2 = CHICKEN,3 = RABBIT,4 = MOUSE,5 = DOG,\n"
                + "6 = CAT,7 = BIRD,8 = FISH,9 = DUCK,10 = RAT");

try {
    input = keyboard.nextInt();
}
catch(Exception e) {
    System.out.println("Error: invalid input");
again = true;

}
if (input >0 && input <=10)
    again = false;

}


        switch (input) {
            case 1:
                star.setType(CartoonType.FOX);
                break;

            case 2:
                star.setType(CartoonType.CHICKEN);
                break;
            case 3:
                star.setType(CartoonType.RABBIT);
                break;
            case 4:
                star.setType(CartoonType.MOUSE);
                break;
            case 5:
                star.setType(CartoonType.DOG);
                break;
            case 6:
                star.setType(CartoonType.CAT);
                break;
            case 7:
                star.setType(CartoonType.BIRD);
                break;
            case 8:
                star.setType(CartoonType.FISH);
                break;
            case 9:
                star.setType(CartoonType.DUCK);
                break;
            case 10:
                star.setType(CartoonType.RAT);
                break;
        }
        popularityNumber(keyboard, star);
    }

    public static void popularityNumber(Scanner keyboard, CartoonStar star) {
        System.out.println("What is the cartoon popularity number?");
        int popularity = keyboard.nextInt();
        star.setPopularityIndex(popularity);
        System.out.println(star.getName() + star.getType() + star.getPopularityIndex());
    }

}

Answer 1

您的程序将永远运行，因为在nextInt不更改扫描仪状态的情况下进行调用会一次又一次地引发异常：如果用户未输入int，则调用keyboard.nextInt()不会更改扫描仪的外观，因此，当您keyboard.nextInt()在下一次迭代中进行调用时，您将获得一个例外。

您需要添加一些代码来读取用户在处理异常后输入的垃圾，以解决此问题：

try {
     ...
} catch(Exception e) {
    System.out.println("Error: invalid input:" + e.getMessage());
    again = true;
    keyboard.next(); // Ignore whatever is entered
}

注意：在这种情况下nextInt()，您无需依赖异常：您可以调用hasNextInt()，而不是调用，并检查扫描程序是否正在查看整数。