c++ - 它有效，直到我输入第二个输入然后它关闭

Question

 #include <iostream>


using namespace std;

int main()
{   int a;
    int b;
    int sum;
    string ans = "";
    cout << "Input a directive. Upon it finishing it will terminate. Codes are: \n Calc \n Exit \n";
    cin >> ans;
    if(ans == "Calc")
    {
        cout << "Welcome to Calculator! Put in a number. Press Enter to put in number. \n";
        cin >> a;

        cout << "Next number \n";
        cin >> b;
        sum = a + b;

        cout << sum << " Is the amount! \n";
        cout << "Input a directive. Upon it finishing it will terminate. Codes are: \n Calc \n Exit \n";
        ans = "";    
    }

    if(ans != "Calc")
    {    
        cout << "Okay";    
    }
}

这行得通，但如果我不输入 Calc，它什么都不做，但打印出来，但如果我不再输入 Calc，它就会关闭。如果我确实输入了计算器，我可以运行它，当它给出答案时，没问题，然后我按任意键它就关闭了。我是这个论坛/网站的新手，不确定位置是否正确。

Answer 1

好吧，如果你想让程序再次接受用户的输入，那么你应该这样写:

#include <iostream>
using namespace std;
int main()
{   int a;
    int b;
    int sum;
    while(true) 
    {
         string ans = "";
         cout << "Input a directive. Upon it finishing it will terminate. Codes are: \n Calc \n Exit \n";
         cin >> ans;
         if(ans == "Calc")
         {
             cout << "Welcome to Calculator! Put in a number. Press Enter to put in number. \n";
             cin >> a;

             cout << "Next number \n";
             cin >> b;
              sum = a + b;

              cout << sum << " Is the amount! \n";
          }
          else if(ans == "Exit") 
          {
              cout << "Bye!\n";    
              return 0;
          }
          cout << "Okay\n";    
     }
}