java - java中异常时无限循环

Question

这个问题已经有答案了:

Reading input using Scanner causes an infinite loop in Java [duplicate] (4 个回答)

已关闭 7 年前。

我准备了以下代码:

public static void main(String[] args) {
Scanner inChoice = new Scanner(System.in);

while(true)
        {
            try
            {
                System.out.println("--------------------------------------------------------");
                System.out.println("Make your choice :");
                System.out.println("For integer - CHOOSE 1");
                System.out.println("For floating-point numbers - CHOOSE 2");
                intChoice = inChoice.nextInt();
                break;
            }
            catch (InputMismatchException imex)
            {
                System.out.println("You have made a wrong selection. Try again");
                continue;
            }
        }
}

问题是当我选择其他东西而不是整数时(例如“w”)。我的目的是给异常(exception)之后重新选择的机会。但相反，我的代码会无限循环并捕获 block 并给我消息:

"--------------------------------------------------------"
"Make your choice :"
"For integer - CHOOSE 1"
"For floating-point numbers - CHOOSE 2"
"You have made a wrong selection. Try again"


"--------------------------------------------------------"
"Make your choice :"
"For integer - CHOOSE 1"
"For floating-point numbers - CHOOSE 2"
"You have made a wrong selection. Try again"


"--------------------------------------------------------"
"Make your choice :"
"For integer - CHOOSE 1"
"For floating-point numbers - CHOOSE 2"
"You have made a wrong selection. Try again"

结束等等...

它不给我重新选择的机会。有人可以解释一下我做错了什么吗？谢谢

Answer 1

当异常发生时，输入不会被消耗。它留在那里。你需要消耗它，否则它会一直抛出异常。

您只需在异常 block 中添加这一行(当然在继续之前):

inChoice.next();