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

The following prints the digits in order of ascending significance (i.e. units, then tens, etc.):

do {
    int digit = n % 10;
    putchar('0' + digit);
    n /= 10;
} while (n > 0);

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";
}

I have seen many answers, but they all forgot to use do {...} while() loop, which is actually the canonical way to solve this problem and handle 0 properly.

My solution is based on this one by Naveen.

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

std::deque<int> digits;
n = abs(n);
do {
    digits.push_front( n % 10);
    n /= 10;
} while (n>0);

What about floor(log(number))+1?

With n digits and using base b you can express any number up to pow(b,n)-1. So to get the number of digits of a number x in base b you can use the inverse function of exponentiation: base-b logarithm. To deal with non-integer results you can use the floor()+1 trick.

PS: This works for integers, not for numbers with decimals (in that case you should know what's the precision of the type you are using).