Question

#include "stdafx.h"
#include <iostream>
#include <stdexcept>

using namespace std;

inline double multplication(double num1, double num2);
inline double division(double num1, double num2);
inline double addition(double num1, double num2);
inline double substraction(double num1, double num2);


int main()
{
        cout << "Welcome to 'The calculator' just enter a number the     operator '*, /, + and -' then the second number enjoy" << endl;
        char redo = 'r'; double result = 0; int cnt = 0;
        while (redo == 'r' || redo == 'R')
        {
                try
                {
                        if (cnt > 0)
                                cout << "Ok same rules as before" << endl;
                        int num1 = 0, num2 = 0; char inputedOperator = '*';
                        cin >> num1 >> inputedOperator >> num2;

                        switch (inputedOperator)
                        {
                        case('*') :
                                result = multplication(num1, num2);
                                break;
                        case('/') :
                                result = division(num1, num2);
                                break;
                        case('+') :
                                result = addition(num1, num2);
                                break;
                        case('-') :
                                result = substraction(num1, num2);
                                break;

                        default:
                                throw runtime_error("Invalid operator");
                        }
                }
                catch (runtime_error err)
                {
                        cout << err.what() << endl;
                        cout << "Enter y to redo or anykey to exit" << endl;
                        cin >> redo;
                        ++cnt;
                        continue;
                }

                cout << "The answer is " << result << "\nPlease enter y to redo or anykey to exit" << endl;
                cin >> redo;
                ++cnt;
        }

        cout << "Thanks for using" << endl;

        return 0;
}

inline double multplication(double num1, double num2)
{
        return num1 * num2;
}
inline double division(double num1, double num2)
{
        return num1 / num2;
}
inline double addition(double num1, double num2)
{
        return num1 + num2;
}
inline double substraction(double num1, double num2)
{
        return num1 - num2;
}

每当我跑步时，3 * 2＆＃39;作为输入，它工作正常，但对于＆＃39; 3 * d＆＃39;它会在我的代码结尾处循环，并且＃3; 3 d 3＆＃39;很好，因为它只是sais无效的运算符。所以我的问题是为什么它会继续循环以及如何解决它？

Answer 1

加入

if (cin.fail())
{
    cin.clear();
    cin.ignore(numeric_limits<int>::max(), '\n');
}

清除无效输入和缓冲区。在cplusplus.com/forum/beginner/2957 /

上找到了相关信息

程序有效，但当输入无效时，它会保持循环c ++

1 个答案: