Checking cin input stream produces an integer

You can check like this:

int x;
cin >> x;

if (cin.fail()) {
    //Not an int.
}

Furthermore, you can continue to get input until you get an int via:

#include <iostream>



int main() {

    int x;
    std::cin >> x;
    while(std::cin.fail()) {
        std::cout << "Error" << std::endl;
        std::cin.clear();
        std::cin.ignore(256,'\n');
        std::cin >> x;
    }
    std::cout << x << std::endl;

    return 0;
}

EDIT: To address the comment below regarding input like 10abc, one could modify the loop to accept a string as an input. Then check the string for any character not a number and handle that situation accordingly. One needs not clear/ignore the input stream in that situation. Verifying the string is just numbers, convert the string back to an integer. I mean, this was just off the cuff. There might be a better way. This won’t work if you’re accepting floats/doubles (would have to add ‘.’ in the search string).

#include <iostream>
#include <string>

int main() {

    std::string theInput;
    int inputAsInt;

    std::getline(std::cin, theInput);

    while(std::cin.fail() || std::cin.eof() || theInput.find_first_not_of("0123456789") != std::string::npos) {

        std::cout << "Error" << std::endl;

        if( theInput.find_first_not_of("0123456789") == std::string::npos) {
            std::cin.clear();
            std::cin.ignore(256,'\n');
        }

        std::getline(std::cin, theInput);
    }

    std::string::size_type st;
    inputAsInt = std::stoi(theInput,&st);
    std::cout << inputAsInt << std::endl;
    return 0;
}