Question

我想在我的代码

中抛出异常后重新启动程序

    System.out.println("please enter an intger to compute its factorial:");
    BufferedReader bufferedreader = new BufferedReader( new InputStreamReader(System.in)); 
    String number ="";
    try {
        try {
            number = bufferedreader.readLine();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();

        }

    intN=Integer.parseInt(number);
        if (intN > 0) {         // from the command line
            FactorialIter f = 
                new FactorialIter(Math.abs(intN));

        }
    } catch (NumberFormatException e) {
        // TODO Auto-generated catch block
        System.out.println("error you should enter a number");
        throw new MyExceptions("try again please use integer numbers");
        //if possible to restart the main 

    }

所以当用户输入一个字符时，程序会抛出一个异常，然后重新启动就可以了吗？

Answer 1

通常，不应该递归地调用main，尤其是为了重新启动程序。

如果要返回程序中的某个点，请使用循环。以下是您可以执行此操作的一个示例：

boolean done = false;
do {
    done = true;
    try {
         ...
    } catch (NumberFormatException e) {
        System.out.println("error you should enter a number");
        done = false;
    }
} while (!done);

每次异常处理程序将done设置为false时，循环将从头开始继续。

Answer 2

你可以把它放在一个循环中。

 While(isValid != TRUE)
 {
      try
      {
           try {
                 number = bufferedreader.readLine();
           } catch (IOException e) {
               // TODO Auto-generated catch block
                e.printStackTrace();
           }

           intN=Integer.parseInt(number);
           if (intN > 0) {         // from the command line
                  FactorialIter f = 
                         new FactorialIter(Math.abs(intN));
               isValid = TRUE;

           }

      }
      catch (NumberFormatException e)
      {
           System.out.println("error you should enter a number");
           System.out.println("try again please use integer numbers");
           isValid = FALSE;
      }

 }

在抛出异常后调用main进行另一次尝试

2 个答案: