Question

所以我现在正在处理一个项目，这一步要求我写一个程序，询问数字以及何时＆＃34; -1＆＃34;输入后，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来修复此问题，但我确定还有另一个＆＃34;正确的＆＃34;这样做的方式，我想学习如何。

感谢任何帮助。（注意：我只在大约一周学习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break的解决方案，并在添加了我想要的其余功能之后，这是最终产品:(如果有人有关于如何改进我的代码或使其更有效的提示请注释！）< / p>

    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

您可以在输入-1后立即终止循环，更改

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

类似

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

，-1分配给numbertyped时，系统不会输入循环体。

根据您的修改，我建议您可以将sum = sum + numbertyped;缩短为sum += numbertyped;，并且可以通过汇总howmany和evencounter来计算oddcounter。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;

Answer 2

在类似的情况下，类似C语言的用户通常不会使用“退出循环”（break）。你需要这样的东西：

    while (true)
    {

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

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

        sum = sum + numbertyped;
    }

作为旁注，“永久循环”可以用for(;;)编写，有些人说这比while(true)好。但是，大多数人认为while(true)更容易阅读。我个人使用for(;;)。无论哪种方式都可以正确优化，因此您将无法获得运行时差异。

关于for(;;)的参考： while (1) Vs. for (;;) Is there a speed difference?

Answer 3

我认为在这种情况下，你可以颠倒while循环中语句的顺序。

        int numbertyped = 0;
        int sum = 0;

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

        while (numbertyped != -1) {

            sum = sum + numbertyped;

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

        }

第一次循环，numberTyped为0，这是完美的 - 它不会退出循环，并且它实际上不会向总和添加任何东西。在那之后，它不会改变总和，直到你检查它是否为-1。

如何在不使用sum ++的情况下正确计算总和;？

3 个答案: