Count The Number of Odd Digits in the Given Integer Using Command Line arguments

Note: Remove the Red color Highlighted Lines such that it works as a code for Odd digits..

Source Code:

#include<stdio.h>

#include<stdlib.h>

void main(int argc, char const*argv[])

{

 int oddCount = 0, evenCount = 0, InputNum, temp ;

InputNum = atoi(argv[1]);

 while (InputNum> 0)

        {

             temp = InputNum % 10;                         /* separate Last digit from number */

              if (temp % 2 == 1)

                {

                  oddCount++;

                 }

              else

                  {

                  evenCount++;

                  }

                   InputNum /= 10;                      /* remove Last digit from num */

         }

printf("Number of Odd digits : %d \nNumber of Even digits: %d\n", oddCount, evenCount);

}

Output:

Also See : Count No Of Odd and Even Digits 

Thanks and regards,

Tech Bird