C++ Checking for an integer. - c++

Question

New to C++. Having issues correctly looping while handling errors. I am trying to check if user input is an integer, and is positive.
do{
cout << "Please enter an integer.";
cin >> n;
if (cin.good())
{
if (n < 0) {cout << "Negative.";}
else {cout << "Positive.";}
}
else
{
cout << "Not an integer.";
cin.clear();
cin.ignore();
}
}while (!cin.good() || n < 0);
cout << "\ndone.";
When a non-integer is entered, the loop breaks. I feel like I am misunderstanding the inherent usage of cin.clear() and cin.ignore() and the status of cin during this loop. If I remove the cin.ignore(), the loop becomes infinite. Why is this? What can I do to make this into an elegantly functioning loop? Thank you.

Answer 1

In your non-integer branch you are invoking further cin methods so cin.good() gets reset to true.
You could change your code to something like this:
while(1) { // <<< loop "forever"
cout << "Please enter an integer.";
cin >> n;
if (cin.good())
{
if (n < 0) {cout << "Negative.";}
else { cout << "Positive."; break; }
} // ^^^^^ break out of loop only if valid +ve integer
else
{
cout << "Not an integer.";
cin.clear();
cin.ignore(INT_MAX, '\n'); // NB: preferred method for flushing cin
}
}
cout << "\ndone.";
or you can simplify it even further like this:
while (!(cin >> n) || n < 0) // <<< note use of "short circuit" logical operation here
{
cout << "Bad input - try again: ";
cin.clear();
cin.ignore(INT_MAX, '\n'); // NB: preferred method for flushing cin
}
cout << "\ndone.";

Answer 2

int n;
while (!(cin >> n)||n<0)//as long as the number entered is not an int or negative, keep checking
{
cout << "Wrong input. Please, try again: ";
cin.clear();//clear input buffer
}
//only gets executed when you've broken out of the while loop, so n must be an int
cout << "Positive.";
cout << "\ndone.";//finished!
Should do what you want.