find the "string length" of an int

Question

basically I want to return the number of digits in the int -> values like this:

(int)1 => 1
(int)123 => 3
(int)12345678 => 8

I know nothing about C, so please bear with me. I know objective c, but I use ints and floats instead of NSNumbers. I realise I could convert the ints into objective c objects, but this seems faffy, and if I can do it with C I'll know it for the future.

Thanks

Answer 1

use

int d = (value == 0 ? 1 : (int)(log10(value)+1));

Note that this doesnt work for negative numbers, you'll have to use

int d = (value == 0 ? 1 : ((int)(log10(fabs(value))+1) + (value < 0 ? 1 : 0)));

which adds 1 for the minus sign, if value is negative.

Answer 2

Probably much faster than using log or int-to-string conversion and without using any library functions is this:

int nDigits(int i)
{
  if (i < 0) i = -i;
  if (i <         10) return 1;
  if (i <        100) return 2;
  if (i <       1000) return 3;
  if (i <      10000) return 4;
  if (i <     100000) return 5;
  if (i <    1000000) return 6;      
  if (i <   10000000) return 7;
  if (i <  100000000) return 8;
  if (i < 1000000000) return 9;
  return 10;
}

EDIT after Jeff Yates concerns:

For those who worry about int sizes different from 32-bits (similar to pmg's solution but still faster because multiplication is faster than division :-)

#include <limits.h>

#define PO10_LIMIT (INT_MAX/10)


int nDigits(int i)
{
  int n,po10;

  if (i < 0) i = -i;
  n=1;
  po10=10;
  while(i>=po10)
  {
    n++;
    if (po10 > PO10_LIMIT) break;
    po10*=10;
  }
  return n;
}