I am working on this program The problem is that when I hit 5 before 4 to quit I

Question

I am working on this program The problem is that when I hit 5 before 4 to quit I have to go out than back in. When i start the program I need to hit 1 than 2 than 3 than 5 than 4 for the output. Here is the code so far:

#include
#include
#include

using namespace std;

void printDepreciation(double value, int year, double perYear);
void straightLine(double value, double years);
void doubleDeclining(double value, double years);
void sumOfYears(double value, int years);

int main()
{
double value = 0.0; // Orignal value of the item
int years = 0;      // No of years to depreciate
int method = 0;     // The method that user choose

cout << "WELCOME TO JOHN'S DEPRECIATION CALCULATOR" << endl;

while (method != 5)
{
    cout << " Enter a depreciation method " << "(1-SL 2-DDB 3-SYD 4-Quit): ";
    cin >> method;
  
    while (method > 0 && method <= 3 )
    {
      cout << "Original value: ";
      cin >> value;
    
      cout << "Number of years: ";
      cin >> years;
      cout << endl;

      if (method == 1)
      {
        cout << "Straight-Line Method ";
        straightLine(value, years);
      }
      else if (method == 2)
      {
        cout << "Double-Declining-Balance Method ";
        doubleDeclining(value, years);
      }
      else if (method == 3)
      {
        cout << "Sum-of-the-Years'-Digits Method ";
        sumOfYears(value, years);
      }
      cout << " Enter a depreciation method " << "(1-SL 2-DDB 3-SYD 4-Quit): ";
      cin >> method;

     } // end while

       if (method == 4) // put a else in front of if
      {
        cout << "Seeya. Have a nice day! " << endl;
      }
      else if (method >= 5)
      {
        cout << "Invalid entry please Try Again " << endl;
      }
    
} // end while

} // end main

//**************************************************************
// The Function calculate the depreciation and calls the print
// function

void straightLine(double value, double years)
{
double perYear = value / years;
for (int i = 1; i <= years; i++)
{
    value -= perYear;
    printDepreciation(value, i, perYear); //call print function
} // end for
} // end straightLine function

//*************************************************************
// This function calculates depreciation by
// double declining method and calls the print Function

void doubleDeclining(double value, double years)
{
for (int i = 1; i <= years; i++)
{
    double perYear = value * 0.4;
    value -= perYear;
    printDepreciation(value, i, perYear); // calls for print function
} // end for
} // end while

//******************************************************************
// This function calculates depreciation by SumOfYears
// method and calls the print Funtion

void sumOfYears(double value, int years)
{
int yearsFactor = years;
double startingValue = value;
for (int i = 1; i <= years; --yearsFactor, ++i)
{
    double z = 0.0; // variable to add up the years
    for (int j = 0; j <= years; j++)
    {
      z += j;
    }
    double perYear = (yearsFactor / z) * startingValue;
    value -= perYear;
    printDepreciation(value, i, perYear); // calls for print funtion
} // end for
} // end sumOfYears

//*******************************************************************
// This function prints out the output

void printDepreciation(double value, int year, double perYear)
{
cout << "End of year" << setw(3) << year << " Depreciation: "
    << setprecision(2) << fixed << setw(7) << perYear
    << " Current value: " << setw(7) << value << endl;

} // end printDepreciation

Explanation / Answer

instead of   while (method > 0 && method <= 3 )

if (method > 0 && method <= 3 ) then no need of usage of break

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.