c++ - 我不明白这个链接错误

Question

我不明白这个链接错误。我有 2 个类(class):

#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;

和

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;

我也有一个四元数类，我相信细节无关紧要，但是这个类有一个非成员运算符*:

 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;
 }

现在这些行产生链接错误，但为什么呢？

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.

这里是链接错误

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

更新

我发现有 2 个问题与此非常接近，但问题在于差异。

在: question 1和 question 2

但在我的例子中，我需要在 2 个模板类之间进行转换，而不是在同一个类和 2 个实例之间进行转换。我希望这会有所帮助!

Answer 1

尝试确保编译器知道您的友元声明应该是模板特化，而不是全新的非模板函数的声明:

friend Vector3<T> operator* <> (const Quaternion<T>& lhs, const Vector3<T>& rhs);

C++ 常见问题解答中讨论了这个常见错误 here .