Question

我正在试图弄清楚为什么我的程序正在跳过用户输入，所以我可以想出一个解决方案。如果我故意给它输入错误并导致抛出InputMismatchException，它会跳过inputDouble = in.nextDouble();行。这是输出的样子：

Please enter floating point value or -1 to stop
15.7
Please enter floating point value or -1 to stop
15.7
Please enter floating point value or -1 to stop
r
Number format is incorrect please try again
Please enter floating point value or -1 to stop
Number format is incorrect please try again
Total is: 31.4

这是我的代码

import java.util.InputMismatchException;
import java.util.Scanner;

public class AddingNumbers {
    public static void main(final String[] args) {
        Scanner in = new Scanner(System.in);
        double inputDouble = 0;
        double total = 0;
        int tries = 0;
        boolean done = false;

        while (!done) {
            if (tries < 2) {
                try {
                    System.out.println("Please enter floating point value or -1 to stop");
                    inputDouble = in.nextDouble();

                    if (inputDouble != -1) {
                        total = total + inputDouble;
                    } else {
                        System.out.println("Total is: " + total);
                        done = true;
                    }
                } catch (InputMismatchException exception) {
                    System.out.println("Number format is incorrect please try again");
                    tries++;
                }
            } else {
                System.out.println("Total is: " + total);
                done = true;
            }
        }

        in.close();
    }
}

Answer 1

您可以尝试在in.nextLine();

下面放置一个空白inputDouble=in.nextDouble();

Answer 2

扫描仪仍然在缓冲区中输入无效。要清除＆＃34;清除＆＃34;您需要使用in.nextLine();的缓冲区。在in.nextLine();之后使用tries++来解决您的问题。

为什么我的程序在捕获异常后会跳过nextDouble（）？

2 个答案: