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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，但它循环，甚至复制、粘贴和修改了以前工作程序中的适当变量。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());
    }

}
4

1 回答 1

3

您的程序将永远运行，因为nextInt在不更改扫描仪状态的情况下调用将一次又一次地导致异常：如果用户没有输入int，则调用keyboard.nextInt()不会更改扫描仪正在查看的内容，因此当您keyboard.nextInt()在下一次迭代中调用时，你会得到一个例外。

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，您不需要依赖异常：nextInt()您可以调用，而不是调用hasNextInt()，并检查扫描仪是否正在查看整数。

于 2013-11-15T00:45:05.647 回答