CS 3323 Topics in Programming Languages

C++ Assignment 3, Fall 1999
Hierarchy of Vehicles
Due November 3, 1999

The purpose of this assignment is to try out several C++ features in a fairly simple example, particularly features involving inheritance, virtual functions, and polymorphism. You are to do up a hierarchy of classes somewhat similar to the examples of shapes in Chapter 10 of your text. (See the Shape class and derived classes in Figure 10.2 of your text.) Your application should resemble the inventory list of vehicles in a car dealership, with a Vehicle class at the top level and two derived classes Truck and Car below that.

You are to use a separate List class that is supplied below (files list.h and list.cpp).

// list.h: a simple list class
//  Assumes a class Vehicle, and uses pointers to
//    members of that class
#ifndef LIST_H
#define LIST_H

#include "vehicle.h"
const MaxVehicles = 50;
class List {
   Vehicle* list[MaxVehicles];  // an array of pointers to Vehicle
   int num; // the current number of vehicles on the list
public:
   List();  // default constructor
   void insert(Vehicle* ); // insert another vehicle
   void sort_inventory(); // sort into order using to key_inventory()
   void print(); // print the list in its current order
};
#endif

// list.cpp: a simple List class to maintain a list of vehicles
//  Assumes a class Vehicle, and uses pointers to
//    members of that class
//  The List class must have member functions
//        key_inventory(), to determine inventory order for sorting
//        print(), to print the whole list
#include "list.h"
#include <stdlib.h>
#include <string.h>
#include <iostream.h>

List::List()
{
   num = 0;
}

void List::insert(Vehicle* vehicle)
{
   if (num >= MaxVehicles) {
      cout << "Overflow of List of Vehicles" << endl;
      exit(1);
   }
   list[num++] = vehicle;
}

void List::sort_inventory() // sort by inventory number
{
   int i, j;
   for (i = 0; i < num - 1; i++)
      for (j = 0; j < num - 1; j++)
         if (strcmp(list[j]->key_inventory(), list[j+1]->key_inventory() )
          > 0) {
              Vehicle* temp = list[j];
              list[j] = list[j+1];
              list[j+1] = temp;
         }
}

void List::print()
{
   cout << "List of all vehicles:" << endl;
   for (int i = 0; i < num; i++)
      list[i]->print();
}
You should define three classes: Vehicle and derived from Vehicle as public derived classes should be Truck and Car.

Here is code you should use for the Vehicle class (file vehicle.h): 

// vehicle.h -- base class declaration
#ifndef VEHICLE_H
#define VEHICLE_H
class Vehicle {
protected:
   char* inventory_code;
   int delivery_date;
public:
   Vehicle();
   Vehicle(char* code, int date);
   ~Vehicle();
   virtual void print();
   char* key_inventory();
};

#endif
Here the member function key_inventory returns the inventory_code as a string. This member function is used inside List to sort items according to the value returned. The data constructor needs to allocate storage for the string and the destructor needs to delete the string. This version of the print function will print the two data members.

The class Truck should have data members int capacity and char* name. The constructor for Truck has 4 input parameters, and it can call the constructor for Vehicle to initialize the data members inventory_code and delivery_date. The function print in Truck should use the print in Vehicle and then print the additional two members.

The class Car should be similarly derived from Vehicle, with data members char* name (the same as Truck and int mileage (different from Truck.

Creating test data can be done with constructors, as in the sample main program: 

#include <iostream.h>
#include "list.h"
#include "vehicle.h"
#include "truck.h"
#include "car.h"

int main()
{
    List list;
    Vehicle v1 ("F99-2387", 990923);
    list.insert(&v1);
    Vehicle v2 ("S98-4572", 981014);
    list.insert(&v2);
    Vehicle v3 ("F99-4587", 990817);
    list.insert(&v3);
    Truck t1   ("F97-4747", 970810, 200, "Dodge Ram");
    list.insert(&t1);
    Truck t2   ("F96-2122", 960920, 150, "Ford F250");
    list.insert(&t2);
    Car c1     ("F98-6569", 981029, 24,  "Chevy Classic");
    list.insert(&c1);
    Car c2     ("F99-2252", 990817, 28,  "Toyota Camery");
    list.insert(&c2);
    Car c3     ("S97-2343", 970321, 30,  "Mazda Protege");
    list.insert(&c3);
    list.sort_inventory();
    list.print();
}
with sample output below: 
List of all vehicles:
Inventory Code: F96-2122, Delivery Date: 960920
   Truck, capacity: 150, name: Ford F250
Inventory Code: F97-4747, Delivery Date: 970810
   Truck, capacity: 200, name: Dodge Ram
Inventory Code: F98-6569, Delivery Date: 981029
   Car, mileage: 24, name: Chevy Classic
Inventory Code: F99-2252, Delivery Date: 990817
   Car, mileage: 28, name: Toyota Camery
Inventory Code: F99-2387, Delivery Date: 990923
Inventory Code: F99-4587, Delivery Date: 990817
Inventory Code: S97-2343, Delivery Date: 970321
   Car, mileage: 30, name: Mazda Protege

Revision Date: 10/21/99