Question

我试图使用循环开关制作计算器程序。

    double firstNum, secondNum;
    int operation;
    Scanner keyboard = new Scanner(System.in);

    System.out.print("Please enter the first number: ");
    firstNum = keyboard.nextDouble();
    System.out.print("\n");

    System.out.print("Please enter the second number: ");
    secondNum = keyboard.nextDouble();
    System.out.print("\n");

    System.out.println("Press 1 for addition");
    System.out.println("Press 2 for subtraction");
    System.out.println("Press 3 for multiplication");
    System.out.println("Press 4 for division");
    System.out.println("Press 0 to end the program");
    System.out.println("\n");

    System.out.print("Which operation would you like to perform? ");
    operation = keyboard.nextInt();
    System.out.print("\n");


    switch (operation)
    {
        case 1:
            System.out.println(firstNum + " + " + secondNum + " = " + (firstNum + secondNum));
            break;
        case 2:
            System.out.println(firstNum + " - " + secondNum + " = " + (firstNum - secondNum));
            break;
        case 3:
            System.out.println(firstNum + " * " + secondNum + " = " + (firstNum * secondNum));
            break;
        case 4:
            if (secondNum ==0 )
                    {
                        System.out.println("You can't divide by zero.");
                    }
                    else
                    {
                        System.out.println(firstNum + " / " + secondNum + " = " + (firstNum / secondNum));

                    }
            break;
        default:
            System.out.println("You have pressed 0. The program is now over.");
    }

所以，我有这个。如果用户只想计算一件事，它工作正常。但是，我想这样做，以便在得到第一个答案后，再次提示他们进入操作（因此程序循环从＆＃34开始;您想要执行哪个操作？＆＃34;）。我希望循环继续，直到他们按0。

我该怎么做呢？我是新手。

感谢您的帮助。

Answer 1

你需要使用while循环，当你准备好退出时，你会跳出while循环。请记住，switch语句中的任何中断都将应用于switch而不是while循环，因此您可能需要维护一个单独的变量，以确定是否不会中断循环。

例如：

boolean continue = true;

while(continue) {
    System.out.print("Which operation would you like to perform? ");
    operation = keyboard.nextInt();
    System.out.print("\n"); 

    switch (operation)
    {
        case 1:
            System.out.println(firstNum + " + " + secondNum + " = " + (firstNum + secondNum));
            break;
        case 2:
            System.out.println(firstNum + " - " + secondNum + " = " + (firstNum - secondNum));
            break;
        case 3:
            System.out.println(firstNum + " * " + secondNum + " = " + (firstNum * secondNum));
            break;
        case 4:
            if (secondNum ==0 )
                    {
                        System.out.println("You can't divide by zero.");
                    }
                    else
                    {
                        System.out.println(firstNum + " / " + secondNum + " = " + (firstNum / secondNum));

                    }
            break;
        default:
            System.out.println("You have pressed 0. The program is now over.");
            continue = false;
            break;
    }
}

这听起来像是一个家庭作业问题，所以花一些时间来理解vs for for循环等等，不要只接受这个答案并运行它。

Answer 2

最佳解决方案是do-while循环，因为您希望的操作至少发生一次

do
{
    System.out.println("What operation would you like to preform?");
    //do math calculating stuff
    System.out.println("Would you like to do another calculation?");
}

while(scannerScan.nextInt != 0);

当您编写do部分时，将要重复执行的所有内容放在do括号内。在这种情况下，我将SOP和注释放在括号中，因此在nextInt() != 0时SOP将继续打印。这样更容易阅读，并且可以消除程序中任何可能导致其他类型的循环失败的问题，从而保证程序至少运行一次。在那之后，如果它失败了，你就知道你到底在哪里，就在while条件所在的一端。

修改

我在默认情况下检查了一个继续，它应该适合你。这样做的是，如果用户输入一个不可接受的值，则执行默认值并执行continue语句，该语句将直接进入while条件。因此，我建议您重新构建默认情况：

do
{
    int x = 0; 

    switch(x)
    {
        case 1:
            System.out.println("Math ops.");
            break;
        case 2:
            //wonderful math
            break;
        case 3:
            //wonders of code
            break;
        default:
            System.out.println("You entered invalid input. Try again. Enter a non-zero number to continue.");
            continue;
    }
}

while(scannerScan.nextInt() != 0);

如何在用户选择结束程序之前一直提示用户？

2 个答案: