program to count number of digits present in a number.

//program to count number of digits present in a number.

#include<stdio.h>
#include<conio.h>

void main()
{
   int n,rem,count=0;
   printf("enter a number for 'n'\n");
  scanf("%d",&n);

  while(n!=0)
    {

        rem=n%10;

       count++;

        n=n/10;    

   }

printf("total digits =%d",count);

getch();

}

--------------------------------------------------------------------------------------------------

logics in mind:-

>first we enter a number(for 'n').

->If we have number 123 then it has 3 digits and they are 3, 2 and 1 or 1,2 and 3. It means, first we have to get 3 then 2 and then 1 and we have to count them.        

                  ->For this, 

                 ->we have to divide by 10 to get last digit as remainder 

                 ->we also use 'count' to start counting  digits.

                -> to get second digit, we get first 12 and for this , we use 123/10.It is done to get integer                           part  only.

->We repeat this until the value reaches 0. We use loop for this as shown above in the code.     

->at last we display total number of digits stored in variable 'count'.