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