How Can I avoid char input for an int variable?

Question

The program below shows a 'int' value being entered and being output at the same time. However, when I entered a character, it goes into an infinite loop displaying the previous 'int' value entered. How can I avoid a character being entered?

#include<iostream>
using namespace std;

int main(){
int n;

while(n!=0){
            cin>>n;
            cout<<n<<endl;
           }
return 0;
}

Answer 1

Reason for Infinite loop:

cin goes into a failed state and that makes it ignore further calls to it, till the error flag and buffer are reset.

cin.clear();
cin.ignore(100, '\n'); //100 --> asks cin to discard 100 characters from the input stream.

Check if input is numeric:

In your code, even a non-int type gets cast to int anyway. There is no way to check if input is numeric, without taking input into a char array, and calling the isdigit() function on each digit.

The function isdigit() can be used to tell digits and alphabets apart. This function is present in the <cctype> header.

An is_int() function would look like this.

for(int i=0; char[i]!='\0';i++){
    if(!isdigit(str[i]))
    return false;
}
return true;

Answer 2

If you want use user define function you can use the ascii/ansi value to restrict the char input.

48 -57 is the range of the 0 to 9 values

Answer 3

#include <iostream>
#include <climits> // for INT_MAX limits
using namespace std;
int main()
{
    int num;
    cout << "Enter a number.\n";
    cin >> num;
    // input validation
    while (cin.fail())
    {
        cin.clear(); // clear input buffer to restore cin to a usable state
        cin.ignore(INT_MAX, '\n'); // ignore last input
        cout << "You can only enter numbers.\n";
        cout << "Enter a number.\n";
        cin >> num;
    }
}