How to get the digits of a number without converting it to a string/ char array?

Question

How do I get what the digits of a number are in C++ without converting it to strings or character arrays?

Answer 1

You want to some thing like this?

 int n = 0;
    std::cin>>n;

    std::deque<int> digits;
    if(n == 0)
    {
        digits.push_front(0);
        return 0;
    }

    n = abs(n);
    while(n > 0)
    {
        digits.push_front( n % 10);
        n = n /10;
    }
    return 0;

Answer 2

Something like this:

int* GetDigits(int num, int * array, int len) {
  for (int i = 0; i < len && num != 0; i++) {
    array[i] = num % 10;
    num /= 10;
  }
}

The mod 10's will get you the digits. The div 10s will advance the number.

Answer 3

Use a sequence of mod 10 and div 10 operations (whatever the syntax is in C++) to assign the digits one at a time to other variables.

In pseudocode

lsd = number mod 10
number = number div 10
next lsd = number mod 10
number = number div 10

etc...

painful! ... but no strings or character arrays.

Answer 4

A simple solution would be to use the log 10 of a number. It returns the total digits of the number - 1. It could be fixed by using converting the number to an int.

int(log10(number)) + 1

Answer 5

Since everybody is chiming in without knowing the question.
Here is my attempt at futility:

#include <iostream>

template<int D> int getDigit(int val)       {return getDigit<D-1>(val/10);}
template<>      int getDigit<1>(int val)    {return val % 10;}

int main()
{
    std::cout << getDigit<5>(1234567) << "\n";
}

Answer 6

Not as cool as Martin York's answer, but addressing just an arbitrary a problem:

You can print a positive integer greater than zero rather simply with recursion:

#include <stdio.h>
void print(int x)
{
    if (x>0) {
        print(x/10);
        putchar(x%10 + '0');
    }
}

This will print out the least significant digit last.