打印两次

我出于教学目的编写了一个简单的程序，除了打印出名称和您选择的计算答案的部分之外，一切正常。 if 语句似乎执行了两次，就好像它在前进之前向后退了一步。

它将打印出“您想继续吗”，但不会提示用户输入是/否，而是会再次打印出计算的答案，然后询问他们是否愿意继续，除非第二次实际上会等待输入。

提前致谢。

这是我的代码：

import java.util.Scanner;

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

        boolean cont;
        String name, yorn;
        double num1, num2, answer;
        int choice, iterator = 0;

        System.out.print("What is your name? ");
        name = reader.nextLine();

        System.out.print("Please enter a number: ");
        num1 = reader.nextDouble();
        if(num1%2 != 0)
        {
            System.out.println(name + ", the number uou entered, " + num1 + ", is odd");
        }
        else
        {
            System.out.println(name + ", the number uou entered, " + num1 + ", is even");
        }

        System.out.print("Please enter a second number: ");
        num2 = reader.nextDouble();
        if(num2%2 != 0)
        {
            System.out.println(name + ", the second number  you entered, " + num2 + ", is odd");
        }
        else
        {
            System.out.println(name + ", the second number you entered, " + num2 + ", is even");
        }

        System.out.println("1. Add");
        System.out.println("2. Subtract");
        System.out.println("3. Multiply");
        System.out.println("4. Divide");
        System.out.print("Please enter the number for the operation you would like to perform on your numbers: ");
        choice = reader.nextInt();

        cont = true;
        while(cont)
        {

            while(choice != 1 && choice != 2 && choice != 3 && choice != 4)
            {
                System.out.print("The number entered is not one of the options. Please choose one of the operations: ");
                choice = reader.nextInt();
            }

            if(choice == 1)
            {
                answer = num1 + num2;
                System.out.println(name +" the sum of " + num1 + " and " + num2 + " is: " + answer);
            }
            else if(choice == 2)
            {
                answer = num1 - num2;
                System.out.println(name +" the difference between " + num1 + " and " + num2 + " is: " + answer);
            }
            else if(choice == 3)
            {
                answer = num1 * num2;
                System.out.println(name +" the product of " + num1 + " and " + num2 + " is: " + answer);
            }
            else //if(choice == 4)
            {
                answer = num1/num2;
                System.out.println(name +" the quotient of " + num1 + " and " + num2 + " is: " + answer);
            }


            System.out.print("Would you like to do anything else (Y/N)? ");
            yorn = reader.nextLine();

            if(yorn.equals("Y"))
            {
                System.out.println("1. Add");
                System.out.println("2. Subtract");
                System.out.println("3. Multiply");
                System.out.println("4. Divide");
                System.out.print("Please enter the number for the operation you would like to perform on your numbers: ");
                choice = reader.nextInt();

            }
            else if (yorn.equals("N"))
            {
                System.out.println("Thank you for using this prgram. Have a good day");
                cont = false;

            }
        }
    }
}

问题是nextDouble()功能。是的，这会读取双精度值，但它也会读取换行符 (\n) 进入缓冲区。所以下次你打电话的时候nextLine()，它会立即解析换行符，而不是等待您的输入。只需致电另一个reader.nextLine()在再次要求输入之前，请清除换行符的缓冲区。