Checking for letters when reading an integer - c++

Question

I'm making an app that requires the user to input a production order (7 digits long) like this:
int order = 0;
cout << "Insert the order number: ";
cin >> ordem;
How can I prevent the user from entering a letter? Like "I2345G789"?
Doing that, my app just enters an infinite loop. I was thinking to use a function like this:
bool isLetter(int a)
{
string s = to_string(a);
for (int i = 0; i < s.size()-1; i++)
{
if (isdigit(s[i]))
{
return false;
}
else
return true;
}
}
And then:
if (isLetter(order))
{
cout << "Insert only numbers \n";
}
But it doesn't work. Why? And how can I improve the code?
PS: I'm very new to programming, so, sorry for any beginner mistakes.

Answer 1

I guess you have a loop around your code in order to ask for the order number again in case it contains non-digits, for example:
while(...)
{
int order = 0;
cout << "Insert the order number: ";
cin >> order;
}
If you enter something that cannot be parsed into an integer, then the input stream will go into failure mode and that might be the reason why you end up in an infinite loop. In order to overcome your problem in a simple way, you could read a string instead:
string order;
while (true)
{
cout << "Insert the order number: ";
cin >> order;
if (isLetter(order))
cout << "Insert only numbers" << endl;
else
break;
}
The function isLetter() now takes a string and looks like this:
bool isLetter(string s)
{
// Return true if the given string contains at least one letter.
for (size_t i = 0; i < s.size(); i++)
if (!isdigit(s[i]))
return true;
// Return false if there are only digits in the given string.
return false;
}
Please note, that it should be i < s.size() and not i < s.size()-1. And maybe you should rename your function isLetter() to hasLetter(), because that would be a bit more correct.