我最近创建了一个程序，该程序将根据用户输入创建数学问题。通过输入1-4，程序可能会产生问题，或者用户可以通过输入5退出。我遇到的唯一问题是，当我输入字符时，程序会陷入无限循环。我可以使用什么功能检查输入的数字是否为数字，以便显示错误消息？
//CIS180 Assignment #4
#include <iostream>
#include <iomanip>
#include <cstdlib>
#include <ctime>
using namespace std;
int main()
{
//Declare variables.
int num1, num2, menuNum;
int addInput, subInput, multInput, divInput;
int addAnswer, subAnswer, multAnswer, divAnswer;
int addSolution, subSolution, multSolution, divSolution;
srand(time(0));
//Display menu.
cout << "Menu" << endl;
cout << "1. Addition problem" << endl;
cout << "2. Subtraction problem" << endl;
cout << "3. Multiplication problem" << endl;
cout << "4. Division problem" << endl;
cout << "5. Quit this program" << endl << endl;
cout << "Enter your choice (1-5): " << endl;
cin >> menuNum;
//Loop that will provide math problems when user inputs number.
while(menuNum != 5)
{
//Check if the input is valid.
while((menuNum < 1) || (menuNum >5))
{
cout << "The valid choices are 1, 2, 3 , 4, and 5. Please choose: " << endl;
cin >> menuNum;
}
//Generate two random numbers for addition and display output.
if(menuNum == 1)
{
num1 = rand()%500 + 1;
num2 = rand()%500 + 1;
addSolution = num1 + num2;
cout << setw(5) << right << num1 << endl;
cout << setw(2) << left << "+" << setw(3) << right << num2 << endl;
cout << setw(5) << fixed << "-----" << endl;
cin >> addAnswer;
//Check if the addition answer input is correct.
if(addAnswer != addSolution)
cout << "Sorry, the correct answer is " << addSolution << "." << endl;
else if(addAnswer == addSolution)
cout << "Congratulations! That's right." << endl << endl;
}
.
.
.
首先，您应该检测输入尝试是否成功：始终在读取后检查读取尝试是否成功。接下来，当您确定无法读取一个值时，您需要使用将流重置为良好状态，clear()并且需要消除任何不良字符，例如，使用ignore()。给定字符通常是输入的，即，用户必须在使用字符之前按回车键，通常可以合理地获得整行。例如：
for (choice = -1; !(1 <= choice && choice <= 5); ) {
if (!(std::cin >> choice)) {
std::cout << "invalid character was added (ignoring the line)\n";
std::cin.clear();
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
}
}
使用std::numeric_limits<std::streamsize>::max()的方法是获取幻数，该幻数将ignore()根据需要生成尽可能多的字符，直到找到具有第二个自变量值的字符为止。
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