[Lief Webster]

Alright, I've got this program that is supposed to ask the user for input of two ordered pairs, then use a class and a + operator to produce the midpoint. Unfortunately, whenever it is run, it gets the ordered pairs and then invariably says the midpoint is
(4370432,4370432). Here is the code:

//Overloading Operators
//Midpoint

#include <iostream>

using namespace std;

class CMidpoint {
	public:
	   int x , y;
	   CMidpoint() {};
	   CMidpoint( int , int );
	   CMidpoint operator + ( CMidpoint );
};

CMidpoint::CMidpoint( int a , int b )
{
	x = a;
	y = b;
}

CMidpoint CMidpoint::operator + ( CMidpoint param )
{
	CMidpoint temp;
	
	temp.x = x + param.x;
	temp.y = y + param.y;

	return (temp);
}

int main()
{
	CMidpoint c;

	int i , j , k , l;

	cout << "Please enter the x and y value (separated by a space) of an ordered pair: " << endl;
	cin >> i >> j;
	cin.ignore();

	CMidpoint a ( i , j );

	cout << "Please enter the x and y value (separated by a space) of another ordered pair: " << endl;
	cin >> k >> l;

	CMidpoint b ( k , l );

	cout << "Your ordered pairs:" << endl;
	cout << "\t(" << i << "," << j << ")" << endl;
	cout << "\t(" << k << "," << l << ")" << endl;

	cout << "(Press any key to confirm...)" << endl;
	cin.get();
	cin.ignore();
	
	a + b = c;

	cout << "The midpoint of the points is ";
	cout << "(" << c.x << "," << c.y << ")." << endl;

	cin.get();
	return 0;
}

How can I get it to display the actual midpoint, rather than some random number?

[Salem]

> a + b = c;
Try
c = a + b;

[Lief Webster]

That works! But why should the order have any bearing on what the result is? Is it because I overloaded the + operator, but c++ has a default = operator?

[Salem]

Does a + b = c work for you for when you use say doubles? I doubt it.

lvalue = rvalue_expression
is written into the language. You can't just rearrange that to suit.

[seeguna]

Let us consider the stmt
c= a + b

In operator overloading concept, value next to + operator i.e operand value' b' only passed to Overloading function that why in ur pgm 'b' value is copied to' param' operand......

operand ' a' value is passed automatically ,we no need to pass it........

As per ur stmt,
if we put a+b=c,
Which value get passed? Think over it?

[Lief Webster]

Alright, I've got another problem. In the following code, I want to set m to equal c/2 (to finish off the midpoint formula). Why is this code incorrect?

CMidpoint CMidpoint::operator / ( CMidpoint otherparam )
{
    CMidpoint othertemp;
    
    othertemp.x = otherparam.x / 2;
    othertemp.y = otherparam.y / 2;
    
    return ( othertemp );
}

m = 2 / c;

[Lief Webster]

Actually, the second code block should say

m = c / 2;

[Salem]

I think you also need a CMidpoint::operator / which takes an int as a parameter, and you use that parameter in your division.

[seeguna]

if u use c=a/b(for overloading /) means i.e correct

but i think that m=c/2 or m=2/c for overloading is wrong.....

[Lief Webster]

I've got another question. I'm not totally grasping this overloading operators concept, and the book I'm trying to (partially) learn from isn't helping matters. It gives me the following program:

class Vector
{
public:
    int x, y;
    Vector operator + (Vector & OtherVector);
    Vecotr & operator += (Vector & OtherVector);
}

Vector Vector::operator + (Vector & OtherVector)
{
    Vector TempVector;
    TempVector.x = x + OtherVector.x;
    TempVector.y = y + OtherVector.y;
    return TempVector;
}

Vector & Vector::operator += (Vector & OtherVector)
{
    x += OtherVector.x;
    y += OtherVector.y;
    return * this;
}

Vector VectorOne;
Vector VectorTwo;
Vector VectorThree;

VectorOne = VectorTwo + VectorThree;
VectorThree += VectorOne;

Besides from the obvious things like lacking main(), I would appreciate it if someone would explain this code block to me, as I can't really understand the book. What's with all the references? Also, this is the first time I've seen 'this' - could someone tell me what that's all about also?

Thanks alot.