Armstrong Number Checking

What is an armstrong number ?

A number n is said to be an armstrong number of order o (o is number of digits in the number n) if it meets following condition.

d₁d₂d₃d₄d₅d₆d₇ d....do = d₁^o + d₂^o + d₃^o + d₄^o + d₅^o + d₆^o + .... d_o^o  = n
where d₁,d2,d3,d4,d₁, ... are digits in a number.
           o is number of digits in a number.
           n is the number that needs to be checked if it is an armstrong number.

Example :
1 = 1¹ = 1
2 = 2¹ = 2
.
9 = 2¹ = 9
153 = 1³ + 5³ + 3³  = 1 + 125 + 27 = 153
371 = 3³ + 7³ + 1³  = 27 + 343 + 1 = 371
.
1634 = 1⁴ + 6⁴ + 3⁴  + 4⁴ = 1 + 1296 + 81 + 256 = 1634

Algorithm : 
The steps involved in finding is a number is an armstrong or not are as follows :
1. Find the number of digits of the number - this will be our order, say O.
2. Take the last digit.
3. Calculate power O of this digit - use any function from here.
4. Add this to some predeclared sum.
5. Remove the last digit from the number.
6. Repeat steps 2 to 5 until number becomes 0.
7. Compare sum and the number - if both are same, it is armstrong number.

/*

 * C Program to check if a number is armstrong or not.

 */

// Includes

#include <stdio.h>

// Declarations

/* isArmstrong

 * Description : A function to check if given number is armstrong or not.

 * Parameters  : i_number is a number to be checked for armstrong.

 * Returns     : unsigned short 0 - for non armstrong number,

                                1 - for armstrong number.

 */

unsigned short isArmstrong(int i_number);

// Definitions

int main()

{

    int i_number;

    printf("Enter number to be checkd for armstrong : ");

    scanf("%d",&i_number);

    

    // Check if the number is armstrong

    if (isArmstrong(i_number))

    {

        printf("Entered number %d is armstrong.",i_number);

    }

    else

    {

        printf("Entered number %d is not armstrong.",i_number);

    }

}

unsigned short isArmstrong(int i_number)

{

    // numbers copy, so that when we remove last digit every time,

    // we don't loose the original number.

    int i_numberCopy = i_number;

    // stores the sum of powers if digits of a number.

    unsigned int sum = 0;

    

    unsigned int   i_numberOfDigits;

    unsigned short us_lastDigit;

    

    // Negative numbers are not armstrong numbers.

    if (i_number > 0)

    {

        i_numberOfDigits = getNumberOfDigits(i_number);

        if (i_numberOfDigits > 0)

        {

            while (0 != i_numberCopy)

            {

                // Take last digit

                us_lastDigit = i_numberCopy%10;

                

                // Calculate the power and add it in sum

                sum += bitwisePower(us_lastDigit, i_numberOfDigits);

                

                // remove the last digit

                i_numberCopy /= 10;

            }

            

            // Check if number is same as that of sum of powers of digit.

            if (i_number == sum)

            {

                return 1;

            }

        }

    }

   

    return 0 ;

}