Undefined reference to friend operator

Question

In my code I have 2 classes declared in header "Geometry.h", Vector & Point. Inside Point class, I have following:

class Point {
// other stuff
    friend Vector operator-(const Point& lhs, const Point& rhs);
}

Vector is defined in "Vector.cpp" & Point is defined in "Point.cpp".

My compiler (GCC) complains about this and I don't know why:

undefined reference to `Geometry::operator-(Geometry::Point const&, Geometry::Point const&)'|

Definition of function in "Point.cpp" looks like this:

Vector operator-(const Point& lhs,
             const Point& rhs)
{
    return Vector(lhs.GetX()-rhs.GetX(),lhs.GetY()-rhs.GetY());
}

at the start of the cpp file I have "using Geometry::Point; using Geometry::Vector;" Is that fine with friend functions or may that be reason why it doesnt compile? — Alexander Bily, Aug 21 '14 at 12:07
@AlexanderBily aschepler asked about the *definition* of the operator, i.e. last code block in your question. It should be inside the `Geometry` namespace (i.e. `namespace Geometry { Vector operator-(const Point& ... }`). — BartoszKP, Aug 21 '14 at 12:12
So the friend operator declared in class doesnt belong to that class scope? Thanks for your help I changed it so everything is defined inside namespace instead of "using" and it works. — Alexander Bily, Aug 21 '14 at 12:14
@AlexanderBily That's only information that this particular function is a friend of this class. It doesn't have anything to do with where is this function defined (however, there is a special syntax that allows you to declare friendship & define the function at the same time. I'd not recommend it however). — BartoszKP, Aug 21 '14 at 12:24
As a side note, you could also add `operaror-=` within the class and then use this operaror in your out-of-class `operator-` without the need to make it a friend. — aaragon, Aug 21 '14 at 13:02
Just for information is there any performance overhead for one more function call or does compiler optimizes calls like this away? — Alexander Bily, Aug 21 '14 at 13:08
@aaragon `operator-` can only be implemented in terms of `operator-=` when the result type is the same class, but that's not the case here. — aschepler, Aug 21 '14 at 13:28
Oh that's true! Which makes you wonder if there's a design issue here. As subtracting apples from apples should give you apples. — aaragon, Aug 21 '14 at 13:31

score 2 · Accepted Answer · answered Aug 21 '14 at 12:21

As you have discovered in a discussion, you should put your operator- code in your namespace:

namespace Geometry
{
    ...
    Vector operator-(const Point& lhs, const Point& rhs)
    {
        return Vector(lhs.GetX()-rhs.GetX(),lhs.GetY()-rhs.GetY());
    }
}

There is an alternative way to put it into a namespace:

Vector Geometry::operator-(const Point& lhs, const Point& rhs)
{
    return Vector(lhs.GetX()-rhs.GetX(),lhs.GetY()-rhs.GetY());
}

Undefined reference to friend operator

1 Answers1