如何在遇到不同條件時顯示錯誤消息

Question

自3個月前我學習了C++編程，目前遇到了一些問題。

這是所期望的輸出，我是被分配給期望的輸出：

雖然對於諧波式平均值和幾何平均值是：

H是調和平均數，而G是幾何平均數。

我嘗試了幾種使用while循環或do-while-loop與if-else語句一起實現預期輸出的方法，但是每當我故意輸入錯誤的字母s，負數或小數，程序針對我直接到節目結束，而不要求重新輸入或接下來的操作進一步輸入...

這是我最新的代碼我提出：

#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; 
} 

幫助和變化，這裏不多讚賞。

Answer 1

你可以做一個字符串如下：

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旅館函數從一個std字符串轉換::爲整數。它定義在

#include <string> header 

你需要反正std :: string。如果您在使用getline之前使用cin，請確保調用cin.ignore以覆蓋緩衝區中留下的空白（When and why do I need to use cin.ignore() in C++?）。