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

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 2

First digit (least significant) = num % 10, second digit = floor(num/10)%10, 3rd digit = floor(num/100)%10. etc

Answer 3

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 4

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

if (n == 0)
    puts("0");
else {
    while (n > 0) {
        int digit = n % 10;
        printf("%d\n", digit);
        n /= 10;
    }
}

Answer 5

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 6

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 7

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

With n digits and using base b you can express any number up to: pow(n,b)-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 decimal (in that case you should know what's the precision of the type you are using).

Answer 8

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);

Answer 9

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.