const Point* pt1

**noobcpp** · 06-30-2007

This is my code

Code:

friend Point operator-(const Point*, const Point*);

Code:

Point operator-(const Point* pt1, const Point* pt2){
	int x = pt1->x - pt2->x;
	int y = pt1->y - pt2->y;
	return Point(x, y);
}

yes I am trying to use Point* Instead of Point&
I am getting this Following Error

Code:

oops_member_op_overload.cpp:26: error: 
‘Point operator-(const Point*, const Point*)’ must have an argument of class or enumerated type
oops_member_op_overload.cpp:51: error: 
‘Point operator-(const Point*, const Point*)’ must have an argument of class or enumerated type

**witek25** · 06-30-2007

maybe you could share a little more of you code to get an idea how you declares and uses your Pointer-class?

**anon** · 06-30-2007

You can't overload these operators for pointers, because pointers are built-in types and they already have these operators defined. Substracting one pointer from another gives you their distance.

You can't use operator overloading to change the way how the language fundamentally works. Similarly, you can't make + operator for int-type do something else than addition.

Code:

#include <iostream>

class A
{
};

int main()
{
    A* p = new A[2];
    std::cout << &p[1] - &p[0] << '\n'; //1
    delete [] p;
}

**Darryl** · 06-30-2007

You can use a smartptr or proxy class. Here a an example for demonstration purposes. (meaning very incomplete for general use)

Code:

#include <iostream>

class Point
{public:

     int mx,my;
     Point (int x, int y):mx(x), my(y)
     {}
};

class PointProxy
{
     Point* mpPoint;
public:

     PointProxy(Point* p):mpPoint(p)
     {}

     Point* operator->()
     {
          return mpPoint;
     }
     Point operator-(PointProxy& proxy)
     {
          return Point(mpPoint->mx - proxy->mx, mpPoint->my - proxy->my);
     }
};




int main()
{
     PointProxy A = new Point(10,15);
     PointProxy B = new Point(12,19);

     Point C = A-B;

     //to show that they are treated like Pointers to point
     std::cout << A->mx << std::endl;
     std::cout << B->mx << std::endl;

     //to show operator- worked
     std::cout << C.mx << std::endl;
     std::cout << C.my << std::endl;

    return 0;
}

**noobcpp** · 06-30-2007

I donno what does this mean can you tell me whats this ??

Code:

Point (int x, int y):mx(x), my(y){}

I am confused with the ':'

Code:

I know Point::function(){}
But not Point::function():somethingelse{}

**noobcpp** · 06-30-2007

Originally Posted by anon

You can't overload these operators for pointers, because pointers are built-in types and they already have these operators defined. Substracting one pointer from another gives you their distance.

So while overloading Do I always need to use

Code:

const class_name&

**Darryl** · 06-30-2007

Originally Posted by noobcpp

I donno what does this mean can you tell me whats this ??

Code:

Point (int x, int y):mx(x), my(y){}

I am confused with the ':'

Code:

I know Point::function(){}
But not Point::function():somethingelse{}

it's a way to initialize your member variable on creation
Though not exactly the same think of

Code:

Point (int x, int y):mx(x), my(y){}

as the same as

Code:

Point (int x, int y)
{
     mx = x;
     my = y:
}

except the first way is more efficient and in some ways the only way to initialize some member variables, like const

**noobcpp** · 06-30-2007

Can you show me an example on that ??

**Darryl** · 06-30-2007

Maybe you should check a reference site like

http://www.cprogramming.com/tutorial...lists-c++.html

**ZuK** · 06-30-2007

Originally Posted by noobcpp

I am confused with the ':'

Code:

I know Point::function(){}
But not Point::function():somethingelse{}

It's called an initializer list and works for constructors only.

Some rather stupid example

Code:

class Circle {
public:
    Circle( int X, int Y, int R );
protected:
   const int x;  // they should not be allowed
   const int y;  // to change after construction
   int r;
};

/* this will not work */
//Circle::Circle( int X, int Y, int R ) {
//    x = X; y = Y; r = R;
//}

/* the only way to initialize x and y */
Circle::Circle( int X, int Y, int R ) 
: x(X), y(Y), r(R)
{

}

Kurt

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