Need to add a loop to exit or repeat program

[Nasutperaah]

Hi all,

I have written a C ++ program and now need to add a way to make it exit on command, or start over from the beginning. I have very little programming knowledge; Here is my code:
Thanks in advance for any help.

///////////////////////////////////////////////////////////////////////////////////////////////////                                                                                      //                                                                
//Program Discription, Show name and number of Stock for purchase                                //
//Calculate 100 shares by 63.25 and add .02 cent commission per share.                           //
//Calculate Commission at 39.95 if Subtotal at or over 39.95                                     //
///////////////////////////////////////////////////////////////////////////////////////////////////                                        
//Version Control:                                                                               // 
//Beta ver set up main Algorythm and function,initalize variables                                //
//perform calculations:                                                                          //
//a=1-100 shares, b=cost per share, c=Subtotal(a*b),d=2cent commission rate                      //
//e=Commission rate* number of shares,f=39.95,g=grandTotal(Commission rate Due + Subtotal Sales) //
//Version 2:                                                                                     //
//1/27/07/                                                                                       //
//Modify code to request and accept user input for number of shares wanted, price of each share  //
//and Grand Total of final Transaction. Add user loop-back structer for new data.                //
///////////////////////////////////////////////////////////////////////////////////////////////////
#include <iostream>
#include <math.h>
#include <iomanip>
using namespace std;
int main()
{
 
    int a=0;
    double b=0,c=0,d=0.02,e=0,f=0;
    cout<<fixed<<showpoint;
 
          cout<<"\n Market Shares For Sale! \n"  <<endl;
 
 
        // user enters number of shares to buy
 
        cout<<"\t Cocoa-Cola Stock. Select the number of shares  (1-1000):$  \n"<<flush;
                 cin>>a;
                   if(a<1 ||a>1000){
                      cout<<"\t Not a valid Entry. Must be 1 to 1000.\n"<<flush<<endl;
                      }
            cout<<endl;
        //user enters cost per share
            cout<<"\t What is the Price Per Share:$  \n"<<flush;
                  cin>>b;
                  cout<<endl;
                  cout<<endl;
               //Find amount of Commission due
 
                 cout<<"\t This is the amount of Commission Due at 2cent a share:$  \n"<<(a*d)<<endl;
                   e=a*d;
                  if (e>39.95){
                        cout<<"\t Commission plus Sub-Total=Grand Total.Please Pay this Amount:$ \n"<<(c+e)<<f;
                        }   
                  else if(e<39.95){        //calculate the price of shares by the number of shares 
                     cout<<"\t This is the Total of your purchase=:  \n"<<(a*b)<<c<<endl;
                  }
 
 
return 0;
 
}

[Banfa]

call/look up the function

void exit(int);

[Nasutperaah]

Hi Benfa,


First thanks for the reply.
After some looking/research I'm still not sure this is the answer to my problem.

I found this example and think I can modify it to suit my needs.

int main()

{

char selection = 'y';



while (tolower(select ion) == 'y') // loop while selection is 'y'

{

cout << "Do you want to continue (y/n):\n";

cin >> selection;

if (cin.fail() || !(tolower(selec tion) == 'y' || tolower(selecti on) == 'n'))

{

cout << "You entered an invalid input, program will restart.\n";

selection = 'y'; //make sure we continue the loop

continue; //jump back to the start of the while loop

}

cin.clear(); //clear the error flag

cin.ignore(100, '\n');// toss out the corrupted input buffer, up to 100 chars



if (tolower(select ion) == 'n')

return 0;

cout << "here we add some code to do some stuff\n";



}


What I still need to know is: do I place the modification at the end of my code or before my code? Or does it even matter?

Thank you so much for the help. I needed a wise pair of eyes!

[Nasutperaah]

Hi Benfa,


First thanks for the reply.
After some looking/research I'm still not sure this is the answer to my problem.

I found this example and think I can modify it to suit my needs.

int main()

{

char selection = 'y';



while (tolower(select ion) == 'y') // loop while selection is 'y'

{

cout << "Do you want to continue (y/n):\n";

cin >> selection;

if (cin.fail() || !(tolower(selec tion) == 'y' || tolower(selecti on) == 'n'))

{

cout << "You entered an invalid input, program will restart.\n";

selection = 'y'; //make sure we continue the loop

continue; //jump back to the start of the while loop

}

cin.clear(); //clear the error flag

cin.ignore(100, '\n');// toss out the corrupted input buffer, up to 100 chars



if (tolower(select ion) == 'n')

return 0;

cout << "here we add some code to do some stuff\n";



}


What I still need to know is: do I place the modification at the end of my code or before my code? Or does it even matter?

Thank you so much for the help. I needed a wise pair of eyes!

Benfa,

my program runs through the code once and when y(for Yes) is entered the program will repeat from the beginning, but only once more. How do I get it to
repeat until the condition q(for Quit) is met?

[RedSon]

Benfa,

my program runs through the code once and when y(for Yes) is entered the program will repeat from the beginning, but only once more. How do I get it to
repeat until the condition q(for Quit) is met?

Look closely at the flow of the program after it enters into the loop. Why would entering in 'y' then run your program one more time and quit? What is your code doing after it takes the input from the user?

[Nasutperaah]

Hi Red S.
I am not sure what you mean. If I pick up the block of code and place it in a second if statement it will run another time. What I want to know is there a way to keep the loop going without making some many blocks of the same code?

[Banfa]

I have to say I see no problem with your last posted code, it requests input, makes sure that the input is valid then if the input is 'n' it returns else it runs the following code block.

[RedSon]

Hi Red S.
I am not sure what you mean. If I pick up the block of code and place it in a second if statement it will run another time. What I want to know is there a way to keep the loop going without making some many blocks of the same code?

I apologize, I thought you were trying to get it to stop after you asked the user for input. They way you have it now if the user says no the program will run one more time then the exit condition for the while loop is met and the code exits. I see now that you want it to continue indefinetly until the user says something. So you want your code to look like this.

bool done = false;
while (!done)
{
 
...your code goes here...
 
  cout << "Do you want to quit?";
  cin >> selection;
 
  if (selection == yes)
  {
     done = true;
  }
}
 

[Nasutperaah]

Thank you for getting back to me and seeing what the real problem was.
I was able to fix the problem with this.

I apologize, I thought you were trying to get it to stop after you asked the user for input. They way you have it now if the user says no the program will run one more time then the exit condition for the while loop is met and the code exits. I see now that you want it to continue indefinetly until the user says something. So you want your code to look like this.

bool done = false;
while (!done)
{
 
...your code goes here...
 
  cout << "Do you want to quit?";
  cin >> selection;
 
  if (selection == yes)
  {
     done = true;
  }
}