How to split a number into digits in C++

Question

I know there are similar questions on Stack overflow about this. I already checked them. Here are two points:

1. The number will be an input from the user, so I won't know how many digits the number may actually contain

2. I DON'T want to directly print the digits, I need to do some further process on every digit, so I need a way to save or assign every single digit. So fore example, if the user enters 1027, I'll need 1, 0, 2, 7 returned, and not printed. That's where my problem starts. Here's how I'd write it if it could be printed:

int x;
cin>>x;
while(x != 0)
{
   cout<<x%10<<endl;
   x/=10;
}

Any hints or help is in advance appreciated.

Answer 1

It depends what order you need it in. If you need least-significant digit (rightmost) to most-significant (leftmost), then your solution is almost there

int x = ...
while(x != 0)
{
   int current = x % 10; // get rightmost digit
   x /= 10;
   // process 'current', or store in a container for later processing
}

If you need most-significant (leftmost) to least-significant (rightmost), then you can do this recursively:

void for_each_digit(int input)
{
  // recursive base case
  if (input == 0) { return; };    

  int current = input % 10
  for_each_digit(input / 10); // recurse *first*, then process
  // process 'current', add to container, etc
}

// ...


int x = ...
for_each_digit(x);

Edit: I apparently missed the part about returning a sequence of digits.

Either approach works. If you go right-to-left, you will need to reverse the container first. If you go recursive, you will need to append each value to the container.

Answer 2

Use a std::string:

std::string input;
std::cin >> input;

Now input[i] is the i-th digit. input.size() is the number of digits.

Answer 3

Well you can use vector. It can take variable length input. You need not declare the size beforehand. Learn more about vector here: vector

#include<iostream>
#include<vector>
#include <algorithm> // std::reverse
using namespace std;

int main(void)
{
    vector<int>digits;
    int x;
    cin >> x;

    while(x)
    {
        digits.push_back(x % 10);
        x = x / 10;
    }

    // reversing the order of the elements inside vector "digits" as they are collected from last to first and we want them from first to last.
    reverse(digits.begin(), digits.end());
    
    // Now the vector "digits" contains the digits of the given number. You can access the elements of a vector using their indices in the same way you access the elements of an array.
    for(int i = 0; i < digits.size(); i++) cout << digits[i] << " ";

    return 0;
}

Answer 4

You may try std::vector<int> to store unknown number of integers as shown:

#include <iostream>
#include <vector>

int main(void) {
    std::vector<int> digits;
    std::string s;

    std::cout << "Enter the number: ";
    std::cin >> s;

    size_t len = s.length();

    for (size_t i = 0; i < len; i++) {
        digits.push_back(s[i] - '0');
    }
    
    // Comment next 3 code to stop getting output
    for (size_t i = 0; i < len; i++)
        std::cout << digits[i] << ' ';
    std::cout << std::endl;

    return 0;
}

---

Note: This approach doesn't performs any mathematical operations (i.e. operation of division and getting remainders). It simply stores each integer in a vector using a for loop.