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Using C++ Write two `void` functions.  These functions each take two  integer parameters.  The functions accomplish the following tasks....

Using C++

Write two `void` functions.  These functions each take two  integer parameters.  The functions accomplish the following tasks.

The first function prints out the numbers from the first argument up to and including the number passed as the second argument counting by the first argument (i.e. if the arguments are 2 and 10,

the information below was given:

#include <iostream>

using namespace std;


// function definitions//


// main program

int main() {

  int firstNumber,secondNumber;

  char countType, doAgain;

  // we will do this until the user is done

  do {

    // collect the numbers from the user

    do {

      cout << "Enter a positive integer: ";

      cin >> firstNumber;

      if (firstNumber < 1) {

        cout << "Error! Invalid number." << endl;

      }

    } while (firstNumber < 1);

    do {

      cout << "Enter another positive integer: ";

      cin >> secondNumber;

      if (secondNumber < 1) {

        cout << "Error! Invalid number." << endl;

      }

    } while (secondNumber < 1);


    // ask for the type of sum

    do {

      cout << "What should I do with these two numbers?" << endl;

      cout << "  - (s)kip counting" << endl;

      cout << "  - (f)actor detection" << endl;

      cout << "Enter choice: ";

      cin >> countType;

      cout << endl;

      if (countType != 's' && countType != 'f') {

        cout << "Error! Invalid selection." << endl;

      }

    } while (countType != 's' && countType != 'f');

    // Make the function call

    switch (countType) {

      case 's':

        printSkipCount(firstNumber,secondNumber);

        break;

      case 'f':

        printFactor(firstNumber,secondNumber);

        break;

    }

    // should we do this again?

    cout << "Try Again? (y/n): ";

    cin >> doAgain;

  } while (doAgain == 'y');

   

  return 0;

   

}

it prints 2 4 6 8 10).

* The second function prints out a message stating if the first argument is a factor of the second argument.

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Homework Answers

Answer #1

PLEASE LOOK THE SCREENSHOT FOR BETTER UNDERSTANDING OF THE PROGRAM.

SOURCE CODE :

#include <iostream>

using namespace std;

// function definitions//

// printSkipCount FUNCTION

void printSkipCount(int firstNumber, int secondNumber)

{

    // looping from firstNumber to SecondNumber

    // were i is incremented by thr firstNumber in each iteration

    for(int i = firstNumber; i <=secondNumber; i += firstNumber)

    //printing the value of i

        cout << i << " ";

    cout << "\n\n";

}

// printFactor FUNCTION

void printFactor(int firstNumber, int secondNumber)

{

    // if the remainder when dividing the secondNumber by firstNumber

    // is zero then firstNumber is a factor of secondNumber

    if (secondNumber % firstNumber == 0)

    //printing the appropriate message

        cout << firstNumber << " is a factor of " << secondNumber;

    cout << "\n\n";

}


// main program

int main()

{

    int firstNumber, secondNumber;

    char countType, doAgain;

    // we will do this until the user is done

    do

    {

        // collect the numbers from the user

        do

        {

            cout << "Enter a positive integer: ";

            cin >> firstNumber;

            if (firstNumber < 1)

            {

                cout << "Error! Invalid number." << endl;

            }

        } while (firstNumber < 1);

        do

        {

            cout << "Enter another positive integer: ";

            cin >> secondNumber;

            if (secondNumber < 1)

            {

                cout << "Error! Invalid number." << endl;

            }

        } while (secondNumber < 1);

        // ask for the type of sum

        do

        {

            cout << "What should I do with these two numbers?" << endl;

            cout << "  - (s)kip counting" << endl;

            cout << "  - (f)actor detection" << endl;

            cout << "Enter choice: ";

            cin >> countType;

            cout << endl;

            if (countType != 's' && countType != 'f')

            {

                cout << "Error! Invalid selection." << endl;

            }

        } while (countType != 's' && countType != 'f');

        // Make the function call

        switch (countType)

        {

        case 's':

            printSkipCount(firstNumber, secondNumber);

            break;

        case 'f':

            printFactor(firstNumber, secondNumber);

            break;

        }

        // should we do this again?

        cout << "Try Again? (y/n): ";

        cin >> doAgain;

    } while (doAgain == 'y');

    return 0;

}

1 #include <iostream> using namespace std; 2 4 5 ~ m n N 000 6 7 // function definitions// // printSkipCount FUNCTION void pr

46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 do { cout << Enter another positive integer: ; cin >> secondNumber

OUTPUT

C:\Users\Faheem\Documents\GitHub\HackerRank>cd c:\Users\Faheem\Documents\GitHub\HackerRank\factorchck\ && g++ skipcount.cpp

HOPE ANSWER THE QUESTION

COMMENT BELOW IF HAVING ANY DOUBTS

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