Association Relationship (concept)

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Hello Guys ..

I'm writing an example to understand the concept of association ..

This is very simple code ..

i'm having problem in getName function .. ?? can you please help me i.e how can i get the name of the associated person, well this is just a class conversation, that i'm trying to implement, that would help in my project ..

Also how can i implement 2 way association .. ??!
Any suggestions would be appreciated ..

C++

#include <iostream>
using namespace::std;

class Person
{
public:
	Person()
	{}
	Person(string GF)
	{married = NULL;}
 //   getName(){ return ??; } ??!

	void marry(Person *p) {married = p;}
	void divorse() {married = NULL;}
	void spouse()
	{if(married == NULL) cout << "Not Married.." << endl;
	else
		cout << "Married With -> " << "what ever ? " << endl; // ??!
	}

private:
	Person *married;
};

int main()
{
	Person *p1, *p2;
	p1 = new Person("John");
	p2 = new Person("Helan");
	p1->spouse();

	p1->marry(p2);
	p1->spouse();
	p2->spouse();


	return 0;
}

ok .. updated version HERE:

C++

#include <iostream>
using namespace::std;

class Person
{
public:
	Person()
	{ }
	Person(string Girl)
	{
		name = Girl;
		married = NULL;
	}

	string getName();
	void marry(Person *p);
	void divorce();
	void spouse();
private:
	Person *married;
	string name;
};

string Person::getName()
{
	return name;
}

void Person::marry(Person *p)
{
	if (married == NULL)
	{
		married  = p;
	//	p->married = this; // for 2way association..!!
	}
	
}
void Person::spouse()
{
	if (married == NULL)
		cout << "Not Married " << endl;
	else
	cout << "Married To: " << married->getName() << endl; // error here
}
void Person::divorce()
{
	if (married == NULL)
	{
		return;
	}
	else
	{
		married->married = NULL;
		married = NULL;
	}
}


int main()
{
	Person *p1, *p2;
	p1 = new Person("John");
	p2 = new Person("Helan");
	p1->spouse();

	p1->marry(p2);
	p1->spouse();
	p2->spouse();


	return 0;
}

Posted 15-Jan-14 9:55am

Usman Hunjra

Updated 16-Jan-14 18:44pm

v3

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Comments

Usman Hunjra 15-Jan-14 15:57pm

Also I'm unable to determine the return type of the function,
Also there is no need to write the BIG THREE..
As I do not manage a resource here ..
Please have an eye .. :?

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Sergey Alexandrovich Kryukov · Accepted Answer · 2014-01-15T12:59:00

Solution 1

This association is already two-way. Just add:

C++

void marry(Person *p) {
    married = p;
    p->married = this;
}
void divorce() // spelling corrected :-)
{
   if (married == NULL) return;
   married->married = NULL;
   married = NULL;
}

// and of course, if you add the field "name" (could be private)

Person mySpouseName = spouse->name; // will be accessible because
                                       // both objects are of the same class

Of course, you would need to add checks for null and other extra arithmetic like that. In C++ and other languages without garbage collection, association relationship is distinctly different from composition, which is also responsible for life time of the composed objects. So, working with associations, you can assume that some other object is holding your objects and will eventually deallocate them. For example, you could have a container object holding a list of all people.

And you mistakenly call the family relationship a "tree". Despite the common notion of the "family tree" (what a misnomer! apparently, it has some paternalistic roots, where female members were not taken seriously :-)), it is not a tree at all. In a tree, by definition, each person would have only one parent, and may have some children — a tree is a graph without cycles. Family, or native relationship between people and other animals is a more complex graph then the tree.

—SA

Posted 15-Jan-14 12:59pm

Sergey Alexandrovich Kryukov

Updated 15-Jan-14 13:00pm

v2

Comments

Ron Beyer 15-Jan-14 19:18pm

+5!

Sergey Alexandrovich Kryukov 15-Jan-14 23:07pm

Thank you, Ron.
—SA

Peter Leow 15-Jan-14 21:29pm

Excellent explanation, +5!

Sergey Alexandrovich Kryukov 15-Jan-14 23:07pm

Thank you, Peter.
—SA

Usman Hunjra 16-Jan-14 0:58am

Thank u so much sir for Great Explanation. :)
i want to use getName() function in the spouse() function to get the name of the associated person .. ???!
like thi: {if(married == NULL) cout << "Not Married.." << endl;
else
cout << "Married With -> " << getName() << endl; // ??!
}
But unable to do that, (errors)
how can i achieve that ..
Thanks For Your Consideration ... !!

Sergey Alexandrovich Kryukov 16-Jan-14 10:29am

You are welcome.

So, what errors? where is your "getName()" method?

And will you accept my answer formally (green "Accept" button)? Your follow-up questions will be welcome anyway?
—SA

Usman Hunjra 16-Jan-14 13:58pm

sir, how to write get name function, what is the return type .. how can i handle this .. ??

Usman Hunjra 16-Jan-14 3:09am

ohh sorry .. this should be..
else
cout << "Married With: " << married->getName() << endl;

Please help me, to rectify this .. ??!

Sergey Alexandrovich Kryukov 16-Jan-14 14:27pm

I already saw it, it looks correct, but where is your std::string getName() {/* ... */}?
—SA

Usman Hunjra 17-Jan-14 1:18am

ok now add the function ..
but in spouse function, the statement: cout << "Married To: " << married->getName() << endl;
cause the error .. no operator find ?? this is the error..

Usman Hunjra 17-Jan-14 1:40am

ok .. i solve this .. with you help sir .. :)

Sergey Alexandrovich Kryukov 17-Jan-14 19:02pm

Great. Good luck, call again.
—SA