java - 如何在不使用 sum++; 的情况下正确计算总和？

Question

所以我目前正在开发一个项目，这一步要求我编写一个程序，要求输入数字，当输入“-1”时，它将计算除-1之外输入的所有数字的总和。我只是通过在循环后添加 +1 来解决这个问题，但我确信还有另一种“正确”的方法来做到这一点，我想了解如何操作。

感谢任何帮助。 (注:我只学习了大约一周的 Java，所以请 ELI5)

public static void main(String[] args) {
            // program in this project exercises 36.1-36.5
            // actually this is just one program that is split in many parts

            Scanner reader = new Scanner(System.in);

            int numbertyped = 0;
            int sum = 0;

            System.out.println("Type numbers: ");

            while (numbertyped != -1) {

                numbertyped = Integer.parseInt(reader.nextLine());

                sum = sum + numbertyped;

            }
            sum++;
            System.out.println("Thank you and see you later!");
            System.out.println("The sum is " + sum);

编辑:我的程序现已完成。我使用了在 while 循环中添加中断的解决方案，并在添加了我想要的其余功能后，这就是最终产品:(如果有人有关于如何改进我的代码或使其更高效的提示，请发表评论!)

    import java.util.Scanner;

    public class LoopsEndingRemembering {

        public static void main(String[] args) {
            // program in this project exercises 36.1-3


        // actually this is just one program that is split in many parts

        Scanner reader = new Scanner(System.in);

        int numbertyped = 0;
        int sum = 0;
        int howmany = 0;
        int evencounter = 0;
        int oddcounter = 0;

        System.out.println("Type numbers: ");

        while (true) {

            numbertyped = Integer.parseInt(reader.nextLine());

            if (numbertyped == -1) 
            {
                break;
            }

            if (numbertyped % 2 == 0)

            {
                evencounter++;
            } 

            else 

            {
                oddcounter++;
            }

            sum = sum + numbertyped;
            howmany++;

        }

        double average = (double) sum / howmany;

        System.out.println("Thank you and see you later!");
        System.out.println("The sum is " + sum);
        System.out.println("How many numbers: " + howmany);
        System.out.println("Average: " + average);
        System.out.println("Even numbers: " + evencounter);
        System.out.println("Odd numbers: " + oddcounter);

    }
}

Answer 1

一旦输入-1，您就可以终止循环，更改

while (numbertyped != -1) {
    numbertyped = Integer.parseInt(reader.nextLine());

类似于

while ((numbertyped = Integer.parseInt(reader.nextLine())) != -1) {
    // ...

当-1被赋值给numbertyped时，不会进入循环体。

根据您的编辑，我建议您可以将 sum = sum + numbertyped; 缩短为 sum += numbertyped; 并且您可以计算 howmany 通过将 evencounter 和 oddcounter 相加。就像，

System.out.println("Type numbers: ");
while (true) {
    numbertyped = Integer.parseInt(reader.nextLine());
    if (numbertyped == -1) {
        break;
    }
    if (numbertyped % 2 == 0) {
        evencounter++;
    } else {
        oddcounter++;
    }
    sum += numbertyped;
}
int howmany = evencounter + oddcounter;
double average = (double) sum / howmany;