Is there a way to detect if an integer that is inputted is a string?

Question

I made a basic calculator. Everything works as intended, the only problem being is that when num1 or num2 are inputted as a string/letter, everything just breaks and causes an endless loop. Is there a way to detect if an integer/double is inputted as a string? If there is, maybe I can make this calculator a little bit better.

#include <iostream>
#include <cmath>

using namespace std;

int main()
{
    double num1;
    double num2;
    string op;
    string rep;
    bool again1 = true;
    bool again = true;

    while(again == true)
    {
    cout << "Enter first number: ";
    cin >> num1;
    cout << "Operation: ";
    cin >> op;
    cout << "Enter second number: ";
    cin >> num2;

    again = false;
    again1 = true;

    if(op == "+" || op == "add")
    {
        cout << num1 << "+" << num2 << " = " << num1+num2 << endl;
    } else if(op == "-"){
        cout << num1 << "-" << num2 << " = " << num1-num2 << endl;
    } else if(op == "*" || op == "x"){
        cout << num1 << "x" << num2 << " = " << num1*num2 << endl;
    } else if(op == "/"){
        if(num2 == 0){
            cout << "(SYNTAX ERROR)";
        } else {
        cout << num1 << "/" << num2 << " = " << num1/num2 << endl;
        }
    }
    cout << endl;


    while(again1 == true){
    again1 = false;
    cout << "Would you like to try again? Y/N: ";
    cin >> rep;
    if(rep == "Y" || rep == "y")
    {
        again = true;
        cout << "Very well!" << endl;
        cout << endl;
    }
    else if(rep == "N" || rep == "n")
    {
        again = false;
        cout << "Ok, see you later!" << endl;
    } else {
        cout << "Invalid Answer" << endl;
        again1 = true;
        }
    }


    }

    return 0;
}

Answer 1

You can take input in strings, and then you can convert those strings to integer or another data type using any appropriate method according to your need here.

If conversion is not possible it may throw an invalid_argument exception that you can catch and ask user to input valid number.