I don't understand this link error - c++

Question

I don't understand this link error.
I have 2 classes:
#include "Vector3.h"
#include "Quaternion.h"
template<typename T>
class Point3 final
{
public:
constexpr Point3(const Vector3<T>& vec)
: x(vec.x), y(vec.y), z(vec.z)
{}
constexpr operator const Vector3<T>() const
{
// It is the equivalent of Vector3 = Point3 - Origin
return Vector3<T>(x, y, z);
}
constexpr operator Vector3<T>() const
{
// It is the equivalent of Vector3 = Point3 - Origin
return Vector3<T>(x, y, z);
}
T x = T(0);
T y = T(0);
T z = T(0);
friend Vector3<T>;
friend Quaternion<T>;
friend Vector3<T> operator*( const Quaternion<T>& lhs, const Vector3<T>& rhs);
friend Vector3<T> operator*( Vector3<T> lhs, const Vector3<T>& rhs);
};
typedef Point3<Float32> Point3f;
and
template<typename T>
class Vector3 final
{
public:
constexpr Vector3()
{}
constexpr Vector3(T _x, T _y, T _z)
: x(_x), y(_y), z(_z)
{}
T x = T(0);
T y = T(0);
T z = T(0);
};
typedef Vector3<Float32> Vector3f;
I also have a Quaternion class the detail are irrelevant i beleive but this class has a non member operator*:
template<typename T>
Vector3<T> operator*( const Quaternion<T>& lhs, const Vector3<T>& rhs)
{
// nVidia SDK implementation
Vector3<T> qvec(lhs.x, lhs.y, lhs.z);
Vector3<T> uv = cross(qvec, rhs) * T(2.0) * lhs.w; //uv = qvec ^ v;
Vector3<T> uuv = cross(qvec, uv) * T(2.0); //uuv = qvec ^ uv;
return rhs + uv + uuv;
}
Now those line produce a link error, but why?
Math::Point3<Float32> pt = -Math::Point3<Float32>::UNIT_Z;
Math::Vector3<Float32> vec = orientation_*pt; // link error here (orientation is a Quaternion<Float32>)
//Math::Vector3<Float32> vec = orientation_*Math::Vector3<Float32>(pt); // this solve the link error.
Here is the link error
Undefined symbols for architecture x86_64:
Math::operator*(Math::Quaternion<float> const&, Math::Vector3<float> const&), referenced from:
GfxObject::Procedural::BoxGenerator::addToTriangleBuffer(GfxObject::Procedural::TriangleBuffer&) const in ProceduralBoxGenerator.o
Update
I found 2 question that are really close to this but the problem relies in the differences.
in:
question 1 and
question 2
But in my case I need to convert between 2 templates classes instead of the same class but 2 instantions. I hope this will help!

Answer 1

Try making sure the compiler knows your friend declaration is supposed to be a template specialization, not a declaration of a brand new non-template function:
friend Vector3<T> operator* <> (const Quaternion<T>& lhs, const Vector3<T>& rhs);
This common mistake is discussed in the C++ FAQ here.

Answer 2

Starting with an SSCCE:
template<typename T> class Quaternion { };
template<typename T> class Vector3 { };
template<typename T> class Point3 {
public:
operator Vector3<T>() const { return Vector3<T>(); }
friend Vector3<T> operator*( const Quaternion<T>& lhs, const Vector3<T>& rhs);
};
template<typename T>
Vector3<T> operator*(const Quaternion<T>& lhs, const Vector3<T>& rhs) { }
int main() {
Quaternion<float> orientation_;
Point3<float> pt;
Vector3<float> vec = orientation_*pt;
return 0;
}
Compiling that with gcc 4.7 I get the following:
x.cc:6:79: warning: friend declaration ‘Vector3<T> operator*(const Quaternion<T>&,
const Vector3<T>&)’ declares a non-template function
[-Wnon-template-friend]
x.cc:6:79: note: (if this is not what you intended, make sure the function template
has already been declared and add <> after the function name here)
Undefined symbols for architecture x86_64:
"operator*(Quaternion<float> const&, Vector3<float> const&)", referenced from:
_main in ccKgI7Ru.o
This points out the problem: you are declaring a friend function, not a friend template. To fix this, you have to do two things: make sure the template declaration preceedes this friendship line, and add angle brackets:
template<typename T>
Vector3<T> operator*(const Quaternion<T>& lhs, const Vector3<T>& rhs) { }
template<typename T> class Point3 {
public:
operator Vector3<T>() const { return Vector3<T>(); }
friend Vector3<T> operator*<>(const Quaternion<T>& lhs, const Vector3<T>& rhs);
};
This will turn the linker error into a compiler error:
error: no match for ‘operator*’ in ‘orientation_ * pt’
note: candidate is:
note: template<class T> Vector3<T> operator*(const Quaternion<T>&, const Vector3<T>&)
note: template argument deduction/substitution failed:
note: ‘Point3<float>’ is not derived from ‘const Vector3<T>’
Which makes sense: declaring a friend is a statement about who may access what data; it does not help a bit in automatic type conversions for template resolution. So I suggest the following instead:
template<typename T> class Point3 {
public:
operator Vector3<T>() const { return Vector3<T>(); }
friend Vector3<T> operator*(const Quaternion<T>& lhs, const Point3& rhs) {
return lhs * Vector3<T>(rhs);
}
};
This declares a new inline friend operator, which will explicitely do the cast you had intended.

Answer 3

Actually to make it work i have to give a body to the friend declared function in the Point class. I wanted to do like item 46 of Effective C++ (3rd edition) from Scott Meyers. like in this question, link.
The key is that the function inside have to have a body. so the class Point3 now become (I have stripped some useless code like MvG did, thanks.
template<typename T> class Point3 {
public:
operator Vector3<T>() const { return Vector3<T>(); }
friend Vector3<T> operator*( const Quaternion<T>& lhs, const Vector3<T>& rhs)
{
return lhs*rhs;
}
};
This make everything compiles and link.
lhsrhs will call the operator defined in Quaternion.h ( that is included in the header file of Point3).
What i don't understand is why i don't have duplicated symbol.
To me the friend operator* inside the Point3 class and the operator* defined in the Quaternion.h header file are the same signature and yet it compile and links.