C++ Checking for an integer.

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.

Paul R · Accepted Answer · 2013-09-02T09:25:35.223

6

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.";

edited Sep 02 '13 at 09:25

answered Sep 02 '13 at 07:19

Paul R

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This works fantastically, thank you! I didn't even think that my choice of loop was holding me back. I appreciate the help and the explanation. – xavi Sep 02 '13 at 07:28
Wouldn't the second version result in an endless loop once the user entered junk (i.e. something like "abc")? You need to extract (or ignore) the junk before looping (but after the `clear()`). – James Kanze Sep 02 '13 at 09:16
1

@James: good catch - thanks - I've fixed it now (and actually tested it !). – Paul R Sep 02 '13 at 09:25

imulsion · Answer 2 · 2013-09-02T07:24:00.100

3

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.

edited Sep 02 '13 at 07:24

answered Sep 02 '13 at 07:17

imulsion

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now that i examine it, it doesn't even do what he asked to do. if the input is good you should leave only if it's negative... – No Idea For Name Sep 02 '13 at 07:20
@NoIdeaForName sorry, didnt spot that. thanks for the catch :) – imulsion Sep 02 '13 at 07:21
If a non-integer is entered, this loop will become infinite. Does the exact same thing as my loop. – xavi Sep 02 '13 at 07:22
It always breaks after entering an integer. The original code should only break if a negative integer is entered. – Onur Sep 02 '13 at 07:22
I believe it should only break if a *positive* integer (>= 0) is entered. – Paul R Sep 02 '13 at 07:22
`cin.clear()` clears error flags, it doesn't clear the input buffer – M.M Jun 13 '19 at 01:33

C++ Checking for an integer.

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