Question

我从3个月前开始学习C ++编程，目前遇到了一些问题。

这是我分配给期望输出的预期输出：

虽然谐波均值和几何平均值的公式是：

H是调和平均值，而G是几何平均值。

我尝试了使用while循环或do-while-loop与if-else语句一起实现期望输出的几种方法，但每当我故意输入错误的字母s，负数或十进制数时，程序就是指示我直接到程序结束，而没有要求重新输入或进一步输入下一步操作...

这是我制作的最新代码：

#include <iostream>
#include <iomanip>
#include <math.h>
using namespace std;
int main()
{
double H, G, n, f, x;
long double HX = 0, GX = 1;
double TypeIn[1000] = {};


cout << "How many values to type?: ";
cin >> n;

while (!(n > 0) && (n == static_cast <int> (n)) && !(cin >> n));
{
    if (n != static_cast <int> (n))
    {
        cout << "No decimal number please: ";
        cin >> n;
    }

    else if (!(cin >> n))
    {
        cin.clear();
        cin.ignore();
        cout << "INTEGER number ONLY: ";
        cin >> n;
    }

    else if (n <= 0)
    {
        cout << "the number must be integer number more than 0, please retype: ";
        cin >> n;
    }
}


for (int k = 0; k < n; k++)
{
    cout << "Enter number #" << k + 1 << ": ";
    cin >> x;

    while (!(x > 0) && (x == static_cast <int> (x)) && !(cin >> x));
    {
        if (x != static_cast <int> (x))
        {
            cout << "No decimal number please: ";
            cin >> x;
        }

        else if (!(cin >> x))
        {
            cin.clear();
            cin.ignore();
            cout << "INTEGER number ONLY: ";
            cin >> x;
        }

        else if (x <= 0)
        {
            cout << "the number must be integer number more than 0, please retype: ";
            cin >> x;
        }
    }
        TypeIn[k] = x;

    HX += 1 / TypeIn[k];
    GX *= TypeIn[k];
}

H = n / HX;
f = 1 / n;
G = pow(GX, f);

cout << "\nFor data:";
for (int k = 0; k < n; k++)
{
    cout << TypeIn[k];
}
cout << setprecision(5);
cout << "\n\nThe harmonic mean is " << H << endl;
cout << "The geometric mean is " << G << endl;

system("pause");
return 0;
}

这里非常感谢帮助和改变。

Answer 1

您可以使用字符串执行以下操作：

string line;
int yourNumber;

getline(cin, s);

//Cycle through the word and check for characters such as '@' which don't fit any of the specific error messages.
for (int i = 0; i < line.length(); i++){
    if (line[i] != '-' || line[i] != '.' || !(line[i] >= '0' && line[i] <= '9'){
        cout << "Generally invalid input such as 5sda2" << endl;
        break;
    }
}

//This line just checks if '-' exists in the string.
if (line.find('-') != std::string::npos)
    cout << "No negatives" << endl;
else if (line.find('.') != std::string::npos)
    cout << "No decimals" << endl;
else
    yourNumber = std::stoi(line);

std :: stoi函数从std :: string转换为整数。它在

中定义

#include <string> header

你还需要std :: string。如果你在使用getline之前使用cin，请务必调用cin.ignore以便克服缓冲区中留下的空白区域（When and why do I need to use cin.ignore() in C++?）。

遇到不同条件时如何显示错误信息

1 个答案: