Problems with forward declaration - Friend functions and line / point classes - c++

Question

I have a demo program for understanding of friend function. I am stuck up with errors related to forward declaration stuff, I guess.
I have a point class which has x & y co-ordinates. The line class has two objects of point class. Now I have a function in line class which will calculate the slope of the line.
This is my program:
#include <iostream>
using namespace std;
class point
{
int x,y;
public:
point(int,int);
point();
friend float line::slope();
};
point::point(int a, int b)
{
x=a;
y=b;
}
point::point()
{
}
class line
{
point p1,p2;
public:
line(point,point);
float slope();
};
line::line(point p1, point p2)
{
this->p1=p1;
this->p2=p2;
}
float line::slope()
{
float s;
s=((float)p2.y-p1.y)/(p2.x-p1.x);
return s;
}
int main()
{
float sl;
point obj(5,10);
point obj1(4,8);
line obj3(obj,obj1);
sl=obj3.slope();
cout<<"\n slope:"<<sl;
return 0;
}
It is giving me compiler errors with respect to forward declarations due to the following:
When I try to define my line class first, it does not know about the point class. Even if I forward declare the point class, that wont suffice coz to create objects of the point class, the compiler should know the size of the point class and hence the whole class itself. Understood it through explanation in this answer: https://stackoverflow.com/a/5543788
If I define the point class first, it needs to know the friend function slope and hence the class line.
So I tried to provide the forward declaration for the line class and the slope function like this before defining the point class:
class line;
float line::slope();
class point
{
int x,y;
public:
point(int,int);
point();
friend float line::slope();
};
Now this gives me the following errors:
friend1.cpp:5: error: invalid use of incomplete type ‘struct line’
friend1.cpp:4: error: forward declaration of ‘struct line’
friend1.cpp:13: error: invalid use of incomplete type ‘struct line’
friend1.cpp:4: error: forward declaration of ‘struct line’
friend1.cpp: In member function ‘float line::slope()’:
friend1.cpp:9: error: ‘int point::y’ is private
friend1.cpp:43: error: within this context
friend1.cpp:9: error: ‘int point::y’ is private
friend1.cpp:43: error: within this context
friend1.cpp:9: error: ‘int point::x’ is private
friend1.cpp:43: error: within this context
friend1.cpp:9: error: ‘int point::x’ is private
friend1.cpp:43: error: within this context
.3. Next I tried to separate out the point class in point.h and point.cpp and line class in line.h and line.cpp. But still here there is a dependency on each other.
Though this should be possible theoretically, I cannot figure it out how to get it working.
Looking out for answers.
Thanks,
Raj
PS: This program is an effort to demonstrate the usage of friend functions alone. Where friend functions are of two types, this is an effort to deal with the second of this kind:
Friend functions which are independent.
Friend functions which are members of another class.
So, usage of friend classes are ruled out in this case.

Answer 1

Add line as a friend, not just a method:
friend class line;
Other remarks:
separate declarations from the implementations in header and implementation files.
prefer full qualification over using directives (i.e. remove using namespace std; and use std::cout instead.
prefer pass-by-reference for complex types - change line(point,point); to line(const point&, const point&);
EDIT For educational purposes - You can't declare that specific function as friend as the code is now because there's no full definition of the line class. Ergo, the following is the only approach:
class point;
class line
{
point *p1,*p2;
public:
line(point,point);
float slope();
};
class point
{
int x,y;
public:
point(int,int);
point();
friend float line::slope();
};
You forward-declare point and change the point members in line to point* (because point isn't a complete type yet). In point you now have the full definition of the line class, so you can declare the method as friend.
EDIT 2: For this particular scenario, it's not possible using point objects inside line because you'd need the full type. But line would also have to be fully defined in order to declare its member as friend.

Answer 2

Just make Line a friend of Point
class point
{
friend class line;
...
};
There's very little purpose in declaring individual methods as friends.

Answer 3

You can create a helper functor to calculate slope. This allows you to make the method of the functor a friend of the point without involving line.
class point;
class line;
struct slope {
float operator () (const point &, const point &) const;
float operator () (const line &) const;
};
class point {
int x,y;
public:
point(int a,int b) : x(a), y(b) {}
point() {}
friend float slope::operator ()(const point &, const point &) const;
};
class line {
point p1,p2;
public:
line(point a,point b) : p1(a), p2(b) {}
float slope() { return ::slope()(*this); }
friend float slope::operator ()(const line &) const;
};
With the implementations:
float slope::operator () (const point &p1, const point &p2) const {
float s;
s=((float)p2.y-p1.y)/(p2.x-p1.x);
return s;
}
float slope::operator () (const line &l) const {
return (*this)(l.p1, l.p2);
}

Answer 4

Under normal circumstances, I would avoid using friend here at all.
Prefer to add a function to point:
float slope_to(const point& other)
{// Not checked, just translated from your implementation
return ((float)other.y-y)/(other.x-x);
}
And implement:
float line::slope()
{
return p1.slope_to(p2);
}
This way line doesn't care about point's implementation and will not need to be changed when you implement 3D points.
Here is a demonstration of a friend of a free function (1.):
#include <iostream>
class point
{
int x, y;
public:
point(int, int);
friend std::ostream& operator <<(std::ostream& os, const point& p);
};
std::ostream& operator <<(std::ostream& os, const point& p)
{
return os << '(' << p.x << ", " << p.y << ')';
}
point::point(int a, int b)
:x(a)
,y(b)
{
}
int main()
{
point obj(5, 10);
std::cout << "\n point " << obj;
}
Here is an example building on your own code to have a friend of a member function (2.).
#include <iostream>
#include <memory>
using namespace std;
class point;
class line
{
std::auto_ptr<point> p1, p2;
public:
line(point&, point&);
float slope();
};
class point
{
int x, y;
public:
point(int, int);
friend float line::slope();
};
point::point(int a, int b)
:x(a)
,y(b)
{
}
line::line(point& p1, point& p2)
:p1(new point(p1))
,p2(new point(p2))
{
}
float line::slope()
{
return ((float)p2->y - p1->y) / (p2->x - p1->x);
}
int main()
{
point obj(5, 10);
point obj1(4, 8);
line obj3(obj, obj1);
cout << "\n slope:" << obj3.slope();
}
In order to reference point from within line, I use (auto_) pointers and references, since the class is declared at that point but not defined. line::slope() is declared in time for me to reference it as a friend of point. This circular dependency is a terrible code smell and should be avoided.

Answer 5

you refer to the line class in friend float line::slope(); before the class is defined.
just add the line class line; before the definition of the Point class
also change friend float line::slope(); to friend class line;