Count total number of digits from a given positive number without looping in C

Question

How to count total number of digits from a given positive number without looping in C?

Answer 1

For integers, take the log10 of the number, round down, and add one.

TEST:

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

int 
num_digits(unsigned long number)
{
    return (int)(floor(log10((double)number)) + 1.0);
}

int
main(int argc, char **argv)
{
    unsigned long test_numbers[] = {
        1, 9, 10, 99, 100, 999, 1000, 9999, 10000, 99999, 100000, 999999,
        123456789ul,
        999999999ul,
        0
    };

    unsigned long *ptr;
    for(ptr = test_numbers; *ptr; ptr++)
    {
        printf("Number of digits in %lu: %d\n", *ptr, num_digits(*ptr));
    }

    return 0;
}

Output:

Number of digits in 1: 1
Number of digits in 9: 1
Number of digits in 10: 2
Number of digits in 99: 2
Number of digits in 100: 3
Number of digits in 999: 3
Number of digits in 1000: 4
Number of digits in 9999: 4
Number of digits in 10000: 5
Number of digits in 99999: 5
Number of digits in 100000: 6
Number of digits in 999999: 6
Number of digits in 123456789: 9
Number of digits in 999999999: 9

Answer 2

Two solutions for you!

int digit_count(unsigned int n)
{
   if (n < 10)
      return 1;
   else
      return (1 + digit_count(n / 10));
}

and

   unsigned int n = 50;
   int i = 0;

HACK:
   i++;

   if (n < 10)
   {
      printf("Digits: %d\n", i);
   }
   else
   {
      n /= 10;

      goto HACK;
   }

Don't hate me for the last one :(

Answer 3

One possible solutions, assuming 16-bit integer values and logical expressions evaluating to 0 or 1. You could replace the conditions with (u > 99) ? 1 : 0 if you're worried about portability.

int digits( unsigned u)
{
    return 1 + (u > 9) + (u > 99) + (u > 999) + (u > 9999);
}