Number Swapping

/*

 * C Program to swap two numbers.

 */

// Includes

#include <stdio.h>

/*

 * Description : Function to swap 2 integers using call by reference.

 * Parameters  : iPtr_num1, iPtr_num2 - pointer to numbers to be swapped.

 * Returns     : Nothing. Swapped numbers are directly reflected in calling function.

 */

void swap(int *iPtr_num1, int *iPtr_num2)

{

    float temp;

    

    // Check if both are pointing to same location.

    // No need to swap in such scenario.

    if (iPtr_num1 == iPtr_num2)

    {

        return;

    }

    

    // Uses temporary variable to swap 2 numbers.

    temp = *iPtr_num1;

    *iPtr_num1 = *iPtr_num2;

    *iPtr_num2 = temp;

}

/*

 * Description : Function to swap 2 integers using addition and subtraction.

 * Parameters  : iPtr_num1, iPtr_num2 - pointer to numbers to be swapped.

 * Returns     : Nothing. Swapped numbers are directly reflected in calling function.

 */

void swapUsingAddSub(int *iPtr_num1, int *iPtr_num2)

{

    // Check if both are pointing to same location.

    // No need to swap in such scenario.

    if (iPtr_num1 == iPtr_num2)

    {

        return;

    }

    

    // Swap using addition and subtraction method

    *iPtr_num1 = *iPtr_num1 + *iPtr_num2;

    *iPtr_num2 = *iPtr_num1 - *iPtr_num2;

    *iPtr_num1 = *iPtr_num1 - *iPtr_num2;

}

/* swapUsingBitwise

 * Description : Function to swap 2 integers using bitwise operators.

 * Parameters  : iPtr_num1, iPtr_num2 - pointer to numbers to be swapped.

 * Returns     : Nothing. Swapped numbers are directly reflected in calling function.

 */

void swapUsingBitwise(int *iPtr_num1, int *iPtr_num2)

{

    // Check if both are pointing to same location.

    // No need to swap in such scenario.

    if (iPtr_num1 == iPtr_num2)

    {

        return;

    }

    

    // Swap using bitwise operators

    *iPtr_num1 = *iPtr_num1 ^ *iPtr_num2;

    *iPtr_num2 = *iPtr_num1 ^ *iPtr_num2;

    *iPtr_num1 = *iPtr_num1 ^ *iPtr_num2;

}

/* swapUsingMultDiv

 * Description : Function to swap 2 integers using multiplication and division.

 * Parameters  : iPtr_num1, iPtr_num2 - numbers to be swapped.

 * Returns     : Nothing. Swapped numbwes are directly reflected in calling function.

 */

void swapUsingMultDiv(float *iPtr_num1, float *iPtr_num2)

{

    // Check if both are pointing to same location.

    // No need to swap in such scenario.

    if (iPtr_num1 == iPtr_num2)

    {

        return;

    }

    

    // Swap using multiplication and division method

    *iPtr_num1 = (*iPtr_num1) * (*iPtr_num2);

    *iPtr_num2 = (*iPtr_num1) / (*iPtr_num2);

    *iPtr_num1 = (*iPtr_num1) / (*iPtr_num2);

}

int main()

{

    int i_num1, i_num2;

    float f_num1, f_num2;

    

    printf("Enter numbwes to be swapped : ");

    scanf("%d %d",&i_num1, &i_num2);

    

    // Swap 2 numbers and print

    swap(&i_num1, &i_num2);

    printf("Swapped                   (%d,%d)\n",i_num1, i_num2);

    

    swapUsingAddSub(&i_num1, &i_num2);

    printf("Swapped using add-sub     (%d,%d)\n",i_num1, i_num2);

    

    swapUsingBitwise(&i_num1, &i_num2);

    printf("Swapped using bitwise xor (%d,%d)\n",i_num1, i_num2);

    

    // For multiplication-division method, arguments needs to be float

    f_num1 = i_num1;    // implicit type casting will take place.

    f_num2 = i_num2;    // implicit type casting will take place.

    swapUsingMultDiv(&f_num1, &f_num2);

    printf("Swapped using mult-div    (%g,%g)\n",f_num1, f_num2);

}

Check For Number Palindrome

Program Logic

• A palindrome number is the one which is same as the number when the digits in number are reversed.
• We keep a copy of original number.
• Try to create reverse number as follows :
  1. Let us say number is 121. Reverse number is initially 0.
  2. We take the last digit i.e. 1.
     Iteration 1 - last digit = 1
     Iteration 2 - last digit = 2
     Iteration 3 - last digit = 1
  3. Multiply reverse number by 10 shifting current digits to right.
     Iteration 1 - reverse number = 0*10 = 0
     Iteration 2 - reverse number = 1*10 = 10
     Iteration 3 - reverse number = 12*10 = 120
  4. Now add last digit to reverse number.
     Iteration 1 - reverse number = 0 + 1 = 1
     Iteration 2 - reverse number = 10 + 2 = 12
     Iteration 3 - reverse number = 120 + 1
  5. Remove the last digit from original number
     Iteration 1 - Original number 12
     Iteration 2 - Original number 1
     Iteration 3 - Original number 0
  5. Repeat steps 2-4 until original number becomes 0.
• Now check if reverse number and copy of original number are the same. If yes, we have a palindrome number :D

/*

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

 */

// Includes

#include <stdio.h>

// Declarations

/* isPalindrome

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

 * Parameters  : i_number - number which needs to be checked for palindrome.

 * Returns     : unsigned short 0 - if a number is not palindrome.

 *                              1 - if a number is palindrome.

 */

unsigned short isPalindrome(int i_number);

// Definitions

int main()

{

    int i_number;

    printf("Enter numbwe that needs to be checked for palindrome : ");

    scanf("%d",&i_number);

    

    // Check for palindrome.

    if (isPalindrome(i_number))

    {

        printf("Entered number %d is palindrome.\n",i_number);

    }

    else

    {

        printf("Entered number %d is not palindrome.\n",i_number);

    }

}

unsigned short isPalindrome(int i_number)

{

    int i_reverseNumber = 0;     // to store reverse number.

    int i_copyNumber = i_number; // to store number copy for modification.

    int i_lastDigit;

    

    while (i_copyNumber)         // unless number becomes 0

    {

        // Take the last digit.

        i_lastDigit = i_copyNumber %10;

        

        // Shift digits to right by 1 place.

        i_reverseNumber *= 10;

        

        // Add last digit on right

        i_reverseNumber += i_lastDigit;

        

        // Remove last digit from original number.

        i_copyNumber /= 10;

    }

    

    return i_reverseNumber == i_number;

}

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 ;

}

Find Number Of Digits In A Number

/*

 * C Program to find number of digits in a number.

 * Pre-Condition : Program needs to be written without using any library function.

 * Example       : If number is 1234 - output will be 4.

 */

// Include

#include <stdio.h>

// Declarations

/* getNumberOfDigits

 * Description : A function that gives number of digits in a given number.

 * Parameters  : i_number is the number of which number of digits needs to be found.

 * Return      : unsigned int data which is number of digits.

 */

unsigned int getNumberOfDigits(int i_number);

// Definitions

int main()

{

    int i_number;

    printf("Enter number of which number of digits needs to find : ");

    scanf("%d",&i_number);

    printf("Number of digits in %d : %u\n",i_number,getNumberOfDigits(i_number));

}

unsigned int getNumberOfDigits(int i_number)

{

    unsigned int count = 0;

    // As long as i_number does not become 0 - repeat

    while (i_number)

    {

        count++;

        i_number/=10;   // reduce the number by 1 digit

    }

    

    return count;

}