java - Java抛出了一个我认为不应该的异常

Question

System.out.println("\nHow many sticks do you want?");
    while(sticks==0){
        try{
            sticks=startInput.nextInt();
        }catch(InputMismatchException e){
            sticks=0;
            System.out.println("Please enter a valid number.");
        }
        sticks=startInput.nextInt();
    }

我要做什么:
它要求一个整数，但是当有人输入字符时，它应该再次询问而不是崩溃。

Answer 1

第二个sticks=startInput.nextInt();不在try-catch块内，因此，如果您放置更多字符，它将再次失败。由于您自己没有处理异常，因此异常会冒出气泡并最终使您的应用程序崩溃。

编辑:根据您的评论，这取决于。假设您想在用户提供0作为您的问题的答案时/应该关闭应用程序，则可以这样做:

System.out.println("\nHow many sticks do you want?");
    while(sticks >= 0){
        try{
            sticks=startInput.nextInt();
        }catch(InputMismatchException e){
            sticks=0;     //Any value which is not 0 will not break your loop. This will be re-populated when the user will supply the number again.
            System.out.println("Please enter a valid number.");
        }
    }

如果您希望应用程序从您的代码中停止(似乎不太可能):

System.out.println("\nHow many sticks do you want?");
    while(sticks==0){
        try{
            sticks=startInput.nextInt();
        }catch(InputMismatchException e){
            sticks=0;     //The 0 will break your while loop, exiting your application gracefully.
            System.out.println("Please enter a valid number.");
        }
    }

这是可行的:

private static int sticks;
    private static Scanner startInput;

    public static void main(String[] args) {
        sticks = 0;
        startInput = new Scanner(System.in);        
        while (sticks >= 0) {            
            try {
                System.out.println("How many sticks do you want?");
                sticks = Integer.parseInt(startInput.nextLine());
            } catch (NumberFormatException e) {
                sticks = 0;     //Any value which is not 0 will not break your loop. This will be re-populated when the user will supply the number again.
                System.out.println("Please enter a valid number.");
            }
        }
    }