How to find out if string is only numbers, no letters - c++

Question

I have constructed a code that checks if the entered number is a proper divisible of 9 by getting the string length and finding the sum of each digit entered and finding if its perfectly divisible by 9.
The code works well, I just need some error checks and I have two things to try and add to it.
I need to make sure first that the inputted information is only numbers, no letters.
Also I want to find a way to make input limited from 1 to 1000.
I have this bool example that will check if it is in fact a number, but how can I make an if statement for my userNum input to check the boolean?
bool isNumber(const string &line)
{
if (line[0] == '0') return true;
return (atoi(line.c_str()));
}
#include <iostream>
#include <string>
using namespace std;
string userNum;
char userInput;
int sum, i, n;
int main()
{
tryagain:
sum = 0;
cout << "Enter number to test if it is divisible by 9: ";
cin >> userNum;
if (userNum == isNumber) {
cout << "Nah..";
return 0;
// Find sum of digits by seperating
n = userNum.length(); // Gets the sum of the string length in bytes
}
for (i = 0; i < n; i++) // Starting from 0, until num is less than the length of userNum go up one increment
sum = (sum + (userNum[i] - '0'));
if (sum % 9 == 0) {
cout << "Your number, " << userNum << " adds up to " << sum << " which is evenly divisible to 9" << endl << endl;
}
else {
cout << "Your number, " << userNum << " adds up to " << sum << " which IS NOT evenly divisible to 9" << endl << endl;
}
// Restart option begins here.
cout << "Would you like to try another number? (Y/N): ";
cin >> userInput;
if (userInput == 'y' || userInput == 'Y') {
goto tryagain;
cout << endl;
}
else if (userInput == 'n' || userInput == 'N') {
cout << endl;
cout << "Goodbye!" << endl;
}
else {
cout << endl;
cout << "You have entered an unknown input, goodbye!" << endl;
}
system("pause");
return 0;
}

Answer 1

You could do this in 3 ways:
Use a regular expression, with std::regex_match. The regex that you are looking for is ^[0-9]+$ which checks if a string contains only numbers. This is a possible expression but not the absolute one
Use std::stoi which tries to convert a string into an integer. Be sure to surround it with a try-catch statement to ensure that the string can be properly parsed. If you get the exception, then the string is not a number.
As Justin has pointed out in the comments, you could also try using std::isdigit and iterate over the string to check each char.
Rely on std functions and try to avoid writing code by yourself. The second approach is probably what you are looking for because, if the method does not throw an exception, you can check the range of the number.
When you face a problem for which you cannot find a solution, give a look at the list of algorithms on cppreference. Use them as much as you can because they make the code shorter, easier to read and they can give you advantages (some algorithms just need an extra parameter at the end to start a parallel execution of the method!)

Answer 2

for (i = 0; i < n; i++) {
if(userNum[i] < '0' || userNum[i] > '9') {
cout << "Nah..";
return 0;
}
sum = (sum + (userNum[i] - '0'));
}

Answer 3

Given the necessary range-check with the range being 1...1000, you do not gain anything from summing the digits and dividing the sum by 9, instead of just doing the latter on the number itself. So the numerical part could be simple i < 1, i > 1000, i % 9 == 0 checks, where i is the parsed integer.
std::stoi parses a number at the beginning of a string, so while you can use it, it does not replace checking the characters. e.g. 1, 1.2, 1a are all parsed to 1. You need a string which does not start with a number, and only then you get an std::invalid_argument exception. So you will need #Justin's suggestion too, though here I flipped it as "any_of the characters is not a digit:
std::string s;
std::cin >> s;
try {
if (std::any_of(s.begin(), s.end(), [](char c) { return !std::isdigit(c); }))
throw std::invalid_argument(" is not all-digits");
int i = std::stoi(s);
if (i < 1) throw std::invalid_argument(" is less than 1");
if (i > 1000) throw std::invalid_argument(" is more than 1000");
if (i % 9 == 0) std::cout << i << " is divisible by 9" << std::endl;
else std::cout << i << " is not divisible by 9" << std::endl;
}
catch (std::invalid_argument ia) {
std::cout << s << ia.what() << std::endl;
}
catch (std::out_of_range oor) {
std::cout << s << " sure is not between 1 and 1000";
}
(stoi's "own" invalid_argument can not occur here, because of the pre-check for all-digits)
In the case you would want to get a number between 1 and 1010000 inclusive, then yes, you could start doing magics, like checking length of input, various patterns, and summing the digits:
(Example is tuned down to 1010 so it can be tested without getting a headache)
std::string s;
std::cin >> s;
std::string::iterator it = s.begin(), end = s.end();
int len = end - it;
try {
// if(len==0) will not happen because of the cin
if (*it == '0') {
if (len == 1) throw std::invalid_argument(" is less than 1");
throw std::invalid_argument(" starts with 0, multi-digit numbers do not do that");
}
if (len > 11) throw std::invalid_argument(" is too long for not being greater than 10^10");
if (len == 11) {
if (*it != '1'
|| std::any_of(it + 1, end, [](char c) { return c != '0'; }))
throw std::invalid_argument(" is greater than 10^10 or is not all-digits");
// well, here we know that the number is 10^10 (and is not divisible by 9)
// so this could return early
}
int sum = 0;
for (; it != end; ++it) {
if(!std::isdigit(*it)) throw std::invalid_argument(" is not all-digits");
sum += (*it) - '0';
}
if (sum % 9 == 0)std::cout << s << " is divisible by 9" << std::endl;
else std::cout << s << " is not divisible by 9" << std::endl;
}
catch (std::invalid_argument ia) {
std::cout << s << ia.what() << std::endl;
}