将数字存储在数组中，C++

Question

I'm taking the digits from a number, how can I store them in an array?

int main()
{
    int n;
    std::cin>>n;
    while (n > 0)
        {
            int digit = n%10;
            n /= 10;
            std::cout<<digit<<" ";
        }

    return 0;
}

Answer 1

The best approach is proably to use a vector since they can be easily resized at runtime, something like:

int main()
{
    int n;
    std::cin>>n;
    std::vector<int> digitArray;
    while (n > 0)
    {
        int digit = n%10;
        n /= 10;
        std::cout<<digit<<" ";
        digitArray.push_back(digit);
    }
    std::reverse(digitArray.begin(), digitArray.end()); // reverse the order of digits

    return 0;
}

Vectors can be accessed just like C-style arrays: digitArray[0] = the first digit.

Another approach to take might be to use a fixed size array since we know that a 32-bit integer can only be up to 10 digits long, something like:

int main()
{
    int n;
    std::cin>>n;
    int digitArray[10] = {0}; //initialize all elements to 0
    int size = 0;
    while (n > 0)
    {
        int digit = n%10;
        n /= 10;
        std::cout<<digit<<" ";
        digitArray[size] = digit;
        ++size;
    }
    std::reverse(std::begin(digitArray), std::begin(digitArray)+(size-1)); // reverse the order of digits

    return 0;
}

This will leave you with useless elements most of the time, though that probably won't matter here.

Answer 2

First count the number of digits, then create a dynamic array and push the digits from highest to lowest index 0 as you move from lsb to msb:

int main()
{
    int n, count = 0;
    std::cin>>n;
    int m = n;
    while(n > 0)
    {
       n /= 10;
       ++count;
    }
    int *a = new int[count];
    while(m > 0)
    {
       a[--count] = m % 10;
       m /= 10;
    }
    delete[] a;  //don't forget to release the memory.
    return 0;   
}

Answer 3

To do this in c , you must first allocate an array. For a 32 bit int, this could be a fixed array with 10 digits (+ NUL char), eg

int main()
{
    int n;
    std::cin >> n;
    char a[11];
    int i = 0;
    while (n > 0) {
        int digit = n % 10;
        a[i] = digit;
        ++i;
        n /= 10;
        std::cout << digit << " ";
    }

    a[i] = '\0'; // final NUL byte
    // the digits are now stored in reverse order in `a`
    return 0;
}

Answer 4

This approach probably helps you to enter any type of digit into an array. even tou can enter any binary digts in it like 00001111.

#include<iostream>
#include<string>
using namespace std;
int main()
{
int n;          //enter size of your number
cin>>n;
string input;
cin>>input;      //enter your number
int arr[n];
for( int i=0;i<n;i++)
{
    arr[i]=input[i] - '0';
}
    for(int i=0;i<n;i++)
    {
        cout<<arr[i];
    }
}