storing digits of a number in array

Question

How do I store different digits of an integer in any array like 1234 to {1,2,3,4} It can be done using char str[]="1234"; printf("%c",str[0]; but how to do it without using string and in integer itself

Answer 1

Here's a snippet that creates an array of digits and prints them out:

#include <stdio.h>
#include <math.h>
#include <stdlib.h>


// Print digits 1 by 1
void numToDigits (int number, int base) {
  int i;
  int n_digits = (int)ceil(log(number+1) / log(base));
  printf("%d digits\n", n_digits);
  int * digits = calloc(n_digits, sizeof(int));

  for (i=0; i<n_digits; ++i) {
    digits[i] = number % base;
    number /= base;
  }

  // digits[0] is the 1's place, so print them starting from the largest index

  for (i=n_digits-1; i>=0; --i) {
    printf("%d", digits[i]);
  }
  printf("\n");

  free(digits);
}

You'll likely want to modify this, but I think it exposes all the important ideas. Don't forget to add -lm when you compile to include math libraries needed for log and ceil. Also note that the printing code isn't made to work with bases larger than 10.