Get number of digits of a number

Question

I have a number like this: int num = 36729; and I want to get the number of digits that compose the number (in this case 5 digits).

How can I do this?

Answer 1

Can't you do this ?

    int num = 36729;
    num.ToString().Length

Answer 2

Use this formula:

if(num)
  return floor(log10(abs((double) num)) + 1);

return 1;

Answer 3

Hint: use the / and the % operators.

Answer 4

Inefficient, but strangely elegant...

#include <stdio.h>
#include <string.h>

int main(void)
{
    // code to get value
    char str[50];
    sprintf(str, "%d", value);

    printf("The %d has %d digits.\n", value, strlen(str));

    return 0;
}

Answer 5

int findcount(int num) 
{ 
    int count = 0; 
    if(num != 0){
      while(num) { 
          num /= 10; 
          count ++; 
      } 
      return count ; 
    }
    else
      return 1;
} 

Answer 6

For any input other than 0, compute the base-10 logarithm of the absolute value of the input, take the floor of that result and add 1:

int dig;
...
if (input == 0)
  dig = 1;
else
  dig = (int) floor(log10(abs((double) input))) + 1;

0 is a special case and has to be handled separately.