在第二次尝试之前,Switch不接受输入

Question

我正在为一个类的项目工作,并且可以使用一些帮助来弄清楚为什么我的代码的行为方式.赋值是提示用户输入两个整数并选择要对它们执行的算术运算,同时避免"除以零"场景.这是我的代码的一部分:

import java.util.Scanner;

public class Program3 
{

public static void main(String[] args) 
{

    double operandOne;
    double operandTwo;
    char newLine = '\n';
    Scanner input = new Scanner (System.in);

    //"input" is an object which calls for input from the keyboard.

        System.out.print("This program will request values for two operands ");
        System.out.println("and perform an arithmetic operation of your choice on them.");
        System.out.println(newLine + "Please enter a value for the first operand.");
        operandOne = input.nextDouble();

        System.out.println("Thank you.  Please enter a value for the second operand.");
        operandTwo = input.nextDouble();

        // This section explains what's going on to the user and requests input for the operands.

        if ( operandTwo == 0 )
        {

            System.out.print("You will not be able to perform division if the ");
            System.out.println("second operand is zero!");
            System.out.println(newLine + "Please choose an option:");
            System.out.println("Type 1 to select a new value for the second operand.");
            System.out.println("Type 2 to continue with a value of zero.");

            // You can't divide by zero!  Are you SURE you want to use that number?

            int reallyWantZero = input.nextInt();
            do
            {   
                System.out.println(newLine + "You must type either 1 or 2 and press enter.");
                reallyWantZero = input.nextInt();
            }

            while ((reallyWantZero < 1) || (reallyWantZero > 2));
            {
                switch (reallyWantZero)
            {
                case 1:
                    System.out.println(newLine + "Please enter a new value for the second operand.");
                    operandTwo = input.nextDouble();
                break;
                case 2:
                    System.out.println(newLine + "Okay, we will proceed.");
                    break;
            }
            }
        }

当我执行代码时,我得到以下内容:

"请输入第一个操作数的值.

1

谢谢.请输入第二个操作数的值.

0

如果第二个操作数为零,您将无法执行除法!

请选择一个选项:

键入1以为第二个操作数选择新值.键入2继续值为零.

2

您必须键入1或2并按Enter键.

2

好的,我们会继续."

我不明白为什么进入2继续不被第一次接受,但它是第二次被接受.任何人都可以帮助解释发生了什么以及如何纠正它？

提前致谢!

编辑:

我将我的代码修改为:

好的,我理解你的意思,但似乎没有用.我将我的代码修改为:

while ((reallyWantZero < 1) || (reallyWantZero > 2));
                {
                    System.out.println(newLine + "You must type either     1 or 2 and press enter.");
                    reallyWantZero = input.nextInt();
                }               
                switch (reallyWantZero)
                {
                    case 1:

但它的行为方式仍然完全相同.此外,如果我输入一个无效的数字,程序只会挂起,并且根本不会返回任何内容.

但它的行为方式仍然完全相同.此外,如果我输入一个无效的数字,程序只会挂起,并且根本不会返回任何内容.

Answer 1

因为do...while循环总是至少执行一次

do
{   
    System.out.println(newLine + "You must type either 1 or 2 and press enter.");
    reallyWantZero = input.nextInt();
} 
while ((reallyWantZero < 1) || (reallyWantZero > 2));

它在执行循环体之后检查一次条件.因此它总是输入代码的这一部分,这意味着将要求用户输入至少两次.您需要使用while循环,它将在输入之前检查循环的条件.

例如

while ((reallyWantZero < 1) || (reallyWantZero > 2)) {
    System.out.println(newLine + "You must type either 1 or 2 and press enter.");
    reallyWantZero = input.nextInt();
}