如何在遇到不同条件时显示错误消息

Question

自3个月前我学习了C++编程，目前遇到了一些问题。

这是所期望的输出，我是被分配给期望的输出：

虽然对于谐波式平均值和几何平均值是：

H是调和平均数，而G是几何平均数。

我尝试了几种使用while循环或do-while-loop与if-else语句一起实现预期输出的方法，但是每当我故意输入错误的字母s，负数或小数，程序针对我直接到节目结束，而不要求重新输入或接下来的操作进一步输入...

这是我最新的代码我提出：

#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; 
} 

帮助和变化，这里不多赞赏。

Answer 1

你可以做一个字符串如下：

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); 

性病:: Stoi旅馆函数从一个std字符串转换::为整数。它定义在

#include <string> header 

你需要反正std :: string。如果您在使用getline之前使用cin，请确保调用cin.ignore以覆盖缓冲区中留下的空白（When and why do I need to use cin.ignore() in C++?）。