How to count the number of digits in numbers in different bases?

Question

With different bases i.e. 10, 8, 16, etc; I'm trying to count the number of characters in each number.

Example

Number: ABCDEF
Number of digits: 6

Any sample or pointer for C?

Answer 1

Here is the program for the above (tested).

#include <stdio.h>

int main()
{
     int num = 0, choice = 0, ret;
     char no[32];
     char frmt[] = {"dox"};

     printf("Entery your choice < 1:desimal > < 2:octal > < 3:hexadecimal > :  ");
     scanf("%d", &choice);

     printf("\nEnter your number : ");

     /*
        After getting the input from user just storing the input number into char array as a string.
     */
     switch (choice) {
         case 1:
             scanf("%d", &num);
             sprintf(no, "%0d", num);
             break;
         case 2:
              scanf("%o", &num);
              sprintf(no, "%0o", num);
              break;
         case 3:
             scanf("%x", &num);
             sprintf(no, "%0x", num);
             break;
         default:
             printf("Invalid choice...\n");
    }
    /*
        The below print will print the input number in respective format i.e decimal/ octal/ hexadecimal.
        and the number of digit for that number.
   */
    printf(" : contains %d digits.\n", printf("%s", no));

    return 0;
}

If you have any question you can ask me :)