How do I avoid using arrays? - c++

Question

My professor wants us to write a program without using arrays or vectors like this:
Write a program using functions that calculates and prints parking charges for each of the n customers who parked their cars in the garage.
Parking rates:
a parking garage charges a $5.00 minimum fee to park for up to five hours.
the garage charges an additional $0.50 per hour for each hour or part thereof in the excess of five hours
the maximum charge for any given 24hr period is $10.00. Assume that no car parks longer that 24 hours at a time.
You should enter the hours parked for each customer. Your program should print the results in a neat tabular format and should calculate and print the total of your receipts.
The program output should look like this:
car------Hours------Charge
1--------2.00--------$5.00
2--------5.00--------$5.00
3--------5.30--------$5.50
etc.
total: 3---12.30----$15.50
I only managed to get this far:
include <iostream>
include <conio.h>
include <cmath>
include <iomanip>
using namespace std;
double calculate(double);
int main()
{
double hours,charge;
int finish;
double sumhours;
sumhours=0;
finish=0;
charge=0;
int cars;
cars=0;
do
{
cout<<"Enter the number of hours the vehicle has been parked: "<<endl;
cin>>hours;
cars++;
sumhours+=hours;
finish=cin.get();
if(hours>24)
{
cout<<"enter a time below 24hrs."<<endl;
cars--;
sumhours=sumhours-hours;
}
}
while(finish!=EOF);
double total=calculate(hours);
cout<<total<<": "<<(cars-1)<<": "<<sumhours;
while(!_kbhit());
return 0;
}
double calculate(double time)
{
double calculate=0;
double fees;
if(time<=5)
return 5;
if(time>15)
return 10;
time=ceil(time);
fees=5+(.5*(time-5));
return calculate;
}

Answer 1

Since this is homework, here is an algorithm:
1. Print header.
2. Clear running total variables.
3. While not end of file
3.1 read a record.
3.2 print record contents
3.3 add record field values to running total variables (Hint! Hint!)
3.4. end-while
4. print out running total variables.
You may have to do some additional calculations with the running total variables, especially for averages.
Edit 1: Example of a running total variable
int sum = 0; // This is the running total variable.
const unsigned int QUANTITY = 23;
for (unsigned int i = 0; i < QUANTITY; ++i)
{
cout << "Adding " << i << " to sum.\n";
sum += i;
}
cout << "Sum is: " << sum << "\n";
cout.flush();
In this example, the data 'i' is not stored only used. The sum variable is a running total.
Look for similarities in your assignment.
Edit 2: Example of detecting end of input on cin
char reply = 'n';
while (tolower(reply) != 'y')
{
cout << "Do you want to quit? (y/n)";
cout.flush();
cin >> reply;
cin.ignore(1000, '\n'); // Eat up newline.
}
cout << "Thanks for the answer.\n";
cout.flush();

Answer 2

Since you can't use arrays or vectors, I think you should print the parking data for each car as it's being processed. Pseudocode:
While more cars:
Read data for next car
Calculate cost
Print data
Add to running totals
End while
Print totals

Answer 3

On every iteration, generate the relevant output, but don't stream it to std::cout. Instead, stream it to a std::stringstream object. Then, at the end, stream that object to std::cout. The maths can be done simply by maintaining a running accumulation of the input values.
This, of course, assumes that using a std::stringstream is not considered "cheating" in the context of this homework exercise.

Answer 4

You can try storing your values in a linked list structure instead of an array. Linked lists work great for dynamic storage.
Try this tutorial, http://www.cprogramming.com/tutorial/lesson15.html

Answer 5

My suggestion then is to use a recursive method, the method first accepts input, asks if there is any more input. If there is more input, it then calls itself. If there is no more input, it outputs it's current car and then returns a sum that's added so far in a structure.
The only problem with this method is that it would output entered cars in reverse of input, but it would do so without an array or a file to save to.