Question

我正在编写一个简单的java控制台游戏。我使用扫描仪从控制台读取输入。我试图验证它我要求一个整数，如果输入一个字母，我不会收到错误。我试过这个：

boolean validResponce = false;
int choice = 0;
while (!validResponce)
{
    try
    {
        choice = stdin.nextInt();
        validResponce = true;
    }
    catch (java.util.InputMismatchException ex)
    {
        System.out.println("I did not understand what you said. Try again: ");
    }
}

但它似乎创建了一个无限循环，只是打印出catch块。我做错了什么。

是的，我是Java的新手

Answer 1

nextInt()不会丢弃不匹配的输出;程序将尝试一遍又一遍地阅读，每次都失败。使用hasNextInt()方法确定在调用int之前是否可以阅读nextInt()。

确保当您在InputStream中找到除整数之外的其他内容时，使用nextLine()清除它，因为hasNextInt()也不会丢弃输入，它只会测试下一个标记输入流。

Answer 2

尝试使用

boolean isInValidResponse = true;
//then
while(isInValidResponse){
//makes more sense and is less confusing
    try{
        //let user know you are now asking for a number, don't just leave empty console
        System.out.println("Please enter a number: ");
        String lineEntered = stdin.nextLine(); //as suggested in accepted answer, it will allow you to exit console waiting for more integers from user
        //test if user entered a number in that line
        int number=Integer.parseInt(lineEntered);
        System.out.println("You entered a number: "+number);
        isInValidResponse = false;
    }
//it tries to read integer from input, the exceptions should be either NumberFormatException, IOException or just Exception
    catch (Exception e){
        System.out.println("I did not understand what you said. Try again: ");
    }
}

由于avoiding negative conditionals https://blog.jetbrains.com/idea/2014/09/the-inspection-connection-issue-2/

的共同话题

尝试catch阻塞导致无限循环？

2 个答案: