Struct Communication and Linked Lists

**CaptainMorgan** · 09-10-2006

I suppose this question could also be placed within the C forum, but here goes. Say I have a linked list for Cars that I have in a used car lot, how do I link another linked list containing past Buyers of the Car accordingly? Within the Buyer's list is a member called Name and I want this same Name to be found in the Cars list so I can link the two lists together so that when I pull up information regarding a Car, the past Buyers will be listed, or if I pull up a Buyer, the Cars owned will be shown.

Not looking for a complete answer, more like a conceptual framework. I've constructed independent lists in the past with no problem, but linking the two together so as to create a database simulation is something Im not used to. Am I overlooking something small? ie - basic linked list concepts? or is there something else?

Thank you.
-Captain

**Mario F.** · 09-10-2006

It is presumable to assume that two diferent persons have the same name, so you may want to connect the the two lists through an ID instead. The same goes for cars, if you want each instance of your Cars class to define not a specific brand/model, but an actual car.

Regardless, to answer you question directly, one possible way is to create another linked list that is responsible for displaying/managing the relationships.

Code:

struct CarBuyers {
   Cars* vehicle;
   Buyers* owner;
   CarBuyers* next;
}

**Salem** · 09-10-2006

Whilst lists seem good for specific things, they make certain other things a lot harder.
Say, find all the owners of red cars.
Using lists for one particular kind of query is just premature optimisation.

A list would have a problem with this scenario
A person buys a car, sells it, then buys it back again at some later time. Rare to be sure, but not impossible.

http://en.wikipedia.org/wiki/Entity-relationship_model
Say for example,
"Car"
+ car registration
- colour
- engine size
- current owner

"Owner"
- name
- address
+ drivers licence number
+ car registration

"Transaction"
- date
+ car registration
- seller
- buyer
- price

The + indicates the key field of each entity. Within a given set of entities, this field is unique.
The owner has two key fields to allow owners to own more than one car.

Each entity is just a struct (or class), and the whole database is just three arrays of those structs.

Draw out the diagram, and annotate the relationships with things like

Code:

      owns:n             owned by:1
car <------------------------------> owner

A car is owned by one owner, but a single owner can own many cars.

**Hikaru** · 09-10-2006

Maybe a vector of pointers into the other list?

Code:

#include <iostream>
#include <string>
#include <vector>

namespace Hikaru
{
    using std::vector;
    using std::string;

    class Buyer;

    class Car
    {
    public:
        Car(const string& name): name_(name) {}
        void addBuyer(Buyer* buyer) { history_.push_back(buyer); }
        const vector<Buyer*>& history() { return history_; }
        const string& name() { return name_; }
    private:
        string name_;
        vector<Buyer*> history_;
    };

    class Buyer
    {
    public:
        Buyer(const string& name): name_(name) {}
        void addCar(Car* car) { history_.push_back(car); }
        const vector<Car*>& history() { return history_; }
        const string& name() { return name_; }
    private:
        string name_;
        vector<Car*> history_;
    };

    void setRelation(Car& car, Buyer& buyer)
    {
        car.addBuyer(&buyer);
        buyer.addCar(&car);
    }

    void showRelation(std::ostream& os, Car& car)
    {
        vector<Buyer*>::const_iterator iter = car.history().begin();
        vector<Buyer*>::const_iterator end = car.history().end();

        os << car.name() << "> ";

        while (iter != end)
        {
            os << (*iter)->name() << ' ';
            ++iter;
        }

        os << '\n';
    }

    void showRelation(std::ostream& os, Buyer& buyer)
    {
        vector<Car*>::const_iterator iter = buyer.history().begin();
        vector<Car*>::const_iterator end = buyer.history().end();

        os << buyer.name() << "> ";

        while (iter != end)
        {
            os << (*iter)->name() << ' ';
            ++iter;
        }

        os << '\n';
    }
}

int main()
{
    using std::cout;
    using Hikaru::Car;
    using Hikaru::Buyer;

    std::vector<Car> cars;
    std::vector<Buyer> buyers;

    cars.push_back(Car("Car 1"));
    cars.push_back(Car("Car 2"));
    cars.push_back(Car("Car 3"));
    buyers.push_back(Buyer("Buyer 1"));
    buyers.push_back(Buyer("Buyer 2"));
    buyers.push_back(Buyer("Buyer 2"));

    // set up relationships
    setRelation(cars[0], buyers[2]);
    setRelation(cars[1], buyers[0]);
    setRelation(cars[2], buyers[1]);
    setRelation(cars[1], buyers[2]);

    // show relationships
    showRelation(cout, cars[0]);
    showRelation(cout, buyers[1]);
    showRelation(cout, cars[1]);

    return 0;
}

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