How to get digit count of an int?

[wtf]

I tried the following code but noticed that it is a bit slow. Any suggestions on how to improve it?

int count_digits(const int d)
{
       int place = 1; int mod = 10;
       if (d == 0)
              {
                     return 0;
              }
       else if (d > 0)
              for (int i = 0; i < d; ++i)
                     {
                            if ( (i % mod) == 9)
                                   {
                                          ++place;
                                          mod *= 10;
                                   }
                     }
       else
              for (int i = 0; i > d; --i)
                     {
                            if ( ( i % mod) == 9)
                                   {
                                          ++place;
                                          mod *=10;
                                   }
                     }
       return place;
}

Also there's a small bug in it somewhere.

It produces correct output from 0 through 99, and 110 through 999, and 1010 through 9999 etc. But input 100 through 109 and it will say it has 2 digits.

[Duthomhas]

Scrap that and rethink.

Given any positive whole number, the number of digits are the number of times you can divide by ten before the number becomes zero.

1234 --> 123 --> 12 --> 1 --> 0
1 2 3 4

Given the number zero, the number of digits is one.

0

Given a negative integer, the number of digits is the number of digits when the number is negated.

-473 (negated) --> 473 --> 47 --> 4 --> 0
1 2 3

You don't need anything but pure integer division and a single loop, plus initial tests for zero and negative numbers.

Hope this helps.