How to reject an integer if certain digits are not wanted - c++

Question

If i want an integer input, but if it contains any digits 2 or 3 in it, it will be rejected.
For example entering 23564 will be invalid.
I was thinking of using do-while loops to solve this, but how do I get it to read the individual digits.

Answer 1

Have a look here:
#include <iostream>
using namespace std;
int main() {
int n,num,flag;
do{
cout<<"Enter number";
cin>>n;
num = n;
flag = 0;
while(num>0)
{
if(num%10 == 2 || num%10 ==3)
{
flag = 1;
break;
}
num = num/10;
}
}while(flag==1);
return 0;
}

Answer 2

If you don't care much about performance, you can convert the integer into a string and look for the digits. This is slower than extracting the last digit and comparing it though.
std::string str = std::to_string(number);
auto found2 = str.find('2');
auto found3 = str.find('3');
//number is valid if it doesn't have a 2 or a 3
bool isValid = found2 == std::string::npos && found3 == std::string::npos;

Answer 3

You can get the last digit of a number with something like:
unsigned int number = 23456;
unsigned int digit = number % 10; // gives 6
You can further "shift" all digits right with:
number = number / 10; // gives 2345
In other words, yes, you can do it using the method you propose:
bool has2Or3(unsigned int number) {
// Do until no more digits left.
while (number != 0) {
// Get last digit, return true if 2 or 3.
int digit = number % 10;
if ((digit == 2) || (digit == 3)) {
return true;
}
// Prepare to get next digit.
number = number / 10;
}
// If NO digits were 2 or 3, return false.
return false;
}
Another way would be to simply convert it to a string and use the string functions to see if it contains a specific digit, something such as:
bool has2Or3(unsigned int number) {
// Get number as string (via string stream).
std::stringstream ss;
ss << number;
std::string s = ss.str();
// If contains 2, return true, else continue.
if (s.find("2") != std::string::npos)
return true;
// If contains 3, return true, else return false.
return (s.find("3") != std::string::npos);
}

Answer 4

Instead of reading the input directly into an integer type, read it into a std::string. Then it's simple to check for unwanted digits; if the string passes the test, convert it to an integer:
int get_value() {
std::string bad = "23";
std::string input = "2"; // start with bad input
while (std::find_first_of(input.begin(), input.end(),
bad.begin(), bad.end()) != input.end()) {
std::cout << "Gimme a number: ";
std::cin >> input;
}
return std::stoi(input);
}