c++ - 有没有办法在平均值中不包含负数，当输入负数时你如何终止程序？

Question

对于上次看到我之前帖子的人来说，我很抱歉。它充满了粗心的错误和错别字。这是我的任务:

"Write a program that will enable the user to enter a series of non-negative numbers via an input statement. At the end of the input process, the program will display: the number of odd numbers and their average; the number of even numbers and their average; the total number of numbers entered. Enable the input process to stop by entering a negative value. Make sure that the user is advised of this ending condition."

这是我的代码:

#include <iostream>
#include <iomanip>

using namespace std;

int main()
{
    int number, total1=0, total2=0, count1=0, count2=0;
    do
    {
        cout << "Please enter a number. The program will add up the odd and even ones separately, and average them: ";
        cin >> number;

        if(number % 2 == 0)
        {
            count1++;
            total1+=number;
        }
        else if (number >= 0)
        {
            count2++;
            total2+=number;
        }
    }
    while (number>=0);
        int avg1 = total1/count1;
        int avg2 = total2/count2;
        cout << "The average of your odd numbers are: " << avg1 << endl;
        cout << "The average of your even numbers are " << avg2 << endl;
}

它似乎工作正常，但是当我输入一个负数来终止程序时，它会将它包含在其余的平均数中。有什么建议可以解决这个问题吗？我知道这是可能的，但我没有想到。

Answer 1

你的主循环应该是这样的:

#include <iostream>

for (int n; std::cout << "Enter a number: " && std::cin >> n && n >= 0; )
{
    // process n
}

或者，如果您想发出诊断信息:

for (int n; ; )
{
    std::cout << "Enter a number: ";

    if (!(std::cin >> n)) { std::cout << "Goodbye!\n"; break; }

    if (n < 0) { std::cout << "Non-positve number!\n"; break; }

    // process n
}