/* A solution to COM1100 Quiz #3 (code 3qa7)
 * by Professor Futrelle, 10/22/2000
 */

#include <iostream>
using namespace std;

// function prototypes; adding parameter names not necessary

void doComps(float MULT, float addend, float divis);
void printResult(int result);

int main()
{
  const float MULT = 2.0e2;
  float addend, divis;

  cout << "\nFor computing ((3 + addend) / divisor) * MULT\n"
       << "please enter the addend and divisor: ";
  cin >> addend >> divis;

  doComps(MULT, addend, divis);

  return 0;
}

// function definitions

void doComps(float MULT, float addend, float divis)
{
  int intResult;
  intResult = ((3 + addend) / divis) * MULT;
  printResult(intResult);
}

void printResult(int result)
{
  cout << "\nThe integer value of the result is: " << result << "\n\n";
}

/* running code produces this, with user input of 7.0 and 2.0:

For computing ((3 + addend) / divisor) * MULT
please enter the addend and divisor: 7.0 2.0

The integer value of the result is: 1000

*/