C Language: How to count digits of a number?

Often when you are writing a script, you need to know how many digits will your input have. Function that  I’ll share with you counts digits of an integer (“100″ – 3 digits; “12″- 2 digits).  User inputs number, scripts counts number of digits of input.

 /*function prototypes */
int count_numbers (int num);
/* main function ask for integer, call the
      count_number function and show the result */

int main (void) {
     int number=0, x=0;

    printf("Please enter an integer: ");
    scanf("%d", &number);
    x=count_numbers (number);
    printf("\n%d consist of %d digits!", number, x);

return 0;
/* Getting input from main function, counting until
     input that we get is not equal to zero,
             incrementing counter by one each time */
 int count_numbers ( int num) {

 int count =0;
 if (num == 0) count++;
   while (num !=0) {
return count;


If you have any questions just ask.

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About Anatoly Spektor

My name is Anatoly Spektor (originally Anatolijs Spektors) I am Software and Web Developer. I have worked in Seneca Center for Development of Open Technology on Big Blue Button Add-on - Polling Module. Currently, I am employed as Software Engineer in Red Hat.
This entry was posted in C, Useful Functions and tagged , , , , , . Bookmark the permalink.

20 Responses to C Language: How to count digits of a number?

  1. hasan says:

    Thank you this post… It helped me! Looking forward for your other C functions….

  2. Andy_G says:

    Yo, i got task dunno what to do with it. If you have time you could help. Here it is:

    ” Write a function called check_passwd() that will compare two strings. It takes two strings as parameters, and returns a character. If the two strings are equal then “A” returned, otherwise the first character that is different from the first string will be returned. For example: ”

    10x man!

  3. shadab says:

    but it only works for not more than 5 digits………..

  4. shadab says:

    int main()
    long int number,dgtcnt=0;
    printf(“Enter an integer number: “);

    printf(“Total no. of digits is: %ld”,dgtcnt);

    try this ………………from shadab

  5. shadab says:

    int main()
    long int number,dgtcnt=0;
    printf(“Enter an integer number: “);

    printf(“Total no. of digits is: %ld”,dgtcnt);

    try this ………………from shadab

  6. saurav kumar says:

    what if number start with zero??

    • Good question, thanks for asking!

      This is a simple program that is intended for natural numbers only. In case of numbers that have zeros in front of it, I would suggest to convert “int” to “string” and count characters.



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  11. RushKP says:

    Hey mate, I typed in the program down below to count the digits in a number, as you have done up above, only I used recursion.

    void main()
    int num;
    int counter;

    count(int a)
    if(a%10==0 && a/10==0)

    My question is what shadab above mentioned- this program returns 5 for any number of >=5 digits…
    Meaning “2343214312431″ will show “5″ as the number of digits.
    Ideas on why this happens?

  12. RushKP says:

    Actually its if(a/10==0) up there in the recursion function. There’s no need for the % operation.

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