How to convert an integer to an array of its digits in C++

Question

Suppose I have the integer 1004. I want to store this in the array A with the following pattern:

    A[0]=1
    A[1]=0
    A[2]=0
    A[3]=4

How can I get the value at that index ? How can I do this in C++?

Answer 1

You get the last index of a number by using modulo 10 and then remove that value by dividing the number by 10.

So assume you do this:

1004 % 10 = 4
1004 / 10 = 100

Then repeat that for each digit

Answer 2

Using c++ static memory:

int originalNumber = 1004;
int digitArray[10] = {0};
int idx = 0;

while (originalNumber > 0)
{
    int digit = n % 10;
    originalNumber /= 10;
    digitArray[idx] = digit;
    ++idx;
}

// Reverse the order of the array
std::reverse(std::begin(digitArray), std::begin(digitArray)+(idx-1));

Answer 3

I'm not sure if it's the most efficient way of doing this but it definitely works.

You can enter each digit in the number to the array but from the end of the array to the beginning.

For example, there is an array with the size of 4, so you get the last digit of the number like this: num % 10, and push the digit to the third index of the array.

Code example:

#define SIZE 4

int* numToArray(int num)
{
    int* arr = new int[SIZE]; // assuming you already know the number of digits in the number
    for(int i = SIZE-1; i >= 0; i++)
    {
        arr[i] = num % 10; // Enters the last digit to the array
        num /= 10; // Gets rid of the last digit in the number
    }
    return arr;
}

Answer 4

Instead of ordinary integer array, I suggest you using std::vector instead.

Then, you can have something like the following:

#include <iostream>
#include <vector>

int main() {
    int number = 1004;
    std::vector<int> digits;

    while (number != 0) {
        digits.insert(digits.begin(), number % 10);
        number /= 10;
    }

    for (auto const i : digits) {
        std::cout << i << " "; // 1 0 0 4
    }

    // or by index

    std::cout << std::endl;

    std::cout << "[0]" << digits[0] << std::endl; // 1
    std::cout << "[1]" << digits[1] << std::endl; // 0
    std::cout << "[2]" << digits[2] << std::endl; // 0
    std::cout << "[3]" << digits[3] << std::endl; // 4

    return 0;
}

Demo

Answer 5

Adding to all the existing answers I'd like to propose a more elegant, if probably less efficient approach utilizing the many wonders of the standard library.

#include <string>
#include <vector>
#include <algorithm>
#include <iostream>

int main()
{
    auto number = 1004;
    auto asString = std::to_string(number); // 
    std::vector<int> digits(asString.length());
    std::transform(asString.begin(), asString.end(), digits.begin(), [](char c){return c -'0';});

    for(auto d : digits)
    {
        std::cout << d << ' ';
    }
}