Question

我正在运行一个简单的Calculator应用程序，以学习Java的异常处理。我设置了两个要处理的异常：InputMismatchException和ArithmeticException，用于除以零。

ArithmeticException被处理，并且do-while循环继续。但是在捕获到InputMismatchException之后，执行终止，而不是继续循环。

代码：

do {
  result = 0;
  System.out.println("Enter the expression to calculate: (eg: 4 + 5) \n");
  try {
    num1 = input.nextInt();
    op = input.next().charAt(0);
    num2 = input.nextInt();
    result = a.Calc(num1, op, num2);             //Calc function to calculate
    System.out.println("= " + result);
  } catch (InputMismatchException e) {
     System.err.println("Please space out the expression");
  } catch (ArithmeticException e) {
     System.err.println("Cannot divide by zero");
  }
  System.out.println("Do you want to try again? (Y to retry)");
  OP = input.next().charAt(0);
} while (OP == 'Y' || OP == 'y');

输出：

Enter the expression to calculate: (eg: 4 + 5)
4 / 0
Cannot divide by zero                  //Exception handled
Do you want to try again? (Y to retry)
Y                                      //Do-while continues

Enter the question to calculate: (eg: 4 + 5)
4+5
Please space out the expression        //Exception handled
Do you want to try again? (Y to retry) //Rest of the code is executed
                                       //But the execution also terminates

预期：在InputMismatchException之后继续工作一段时间

这是正确的方法吗？

Answer 1

InputMismatchException是由于调用nextInt()引起的，因为下一个令牌是4+5。

失败的呼叫未使用令牌。

这意味着OP = input.next().charAt(0)设置了OP = '4'，如果您调试代码，这应该很明显。参见What is a debugger and how can it help me diagnose problems?和How to debug small programs。

您需要丢弃失败的令牌，例如通过在nextLine()子句中调用catch：

} catch (InputMismatchException e) {
    System.err.println("Please space out the expression");
    input.nextLine(); // Discard input(s)
} ...

即使捕获到异常，执行也会终止

1 个答案: