CRTP和表达式模板线性代数

Question

我试图修改我的线性代数模块，以避免虚拟vtable的事情。尝试使用CRTP和表达式模板。 我采用了应该测试整个事情的基本方法，但我无法使其正常工作。

我有4个类别，例如：基础CRTP类别，此处为Mathbase

template <typename Derived>
class Mathbase
{
 public:
 using T = typename dense_traits<Derived>::T;

  Derived& derived() { return static_cast<Derived&>(*this); }
  const Derived& derived() const { return static_cast<const Derived&>(*this); }

  T& coeff(std::size_t row, std::size_t col) { return derived().coeff(row, col); }
  const T& coeff(std::size_t row, std::size_t col) const { return derived().coeff(row, col); }
  T& coeff(std::size_t index) { return derived().coeff(index); }
  const T& coeff(std::size_t index) const { return derived().coeff(index); }
};

然后，在其中实现了转置，行列式等功能的Densebase：

template <typename Derived>
class Densebase : public Mathbase<Derived>
{
 public:

 using Submat = Subview<Derived, dense_traits<Derived>::M-1, dense_traits<Derived>::N-1>;
 using ConstSubmat = const Subview<const Derived, dense_traits<Derived>::M-1, dense_traits<Derived>::N-1>;

  Submat sub(std::size_t row, std::size_t col) { return Submat(derived(), row, col); }
  ConstSubmat sub(std::size_t row, std::size_t col) const { return ConstSubmat(derived(), row, col); }
};

请注意，它声明了两种在父矩阵（co-matrix）上进行引用的类型。然后我有了Matrix类，该类通过访问其存储来实现coeff函数：

template <typename T, std::size_t M, std::size_t N>
class Matrix : public Densebase<Matrix<T,M,N>>
{
  // nothing fancy
};

现在，两件事不起作用：

• 我实现了一个使用拉普拉斯展开的行列式帮助程序（计算共矩阵行列式并将其求和，可惜它无法编译

   template <typename Derived, std::size_t N> struct det_helper { static inline typename dense_traits<Derived>::T run(const Densebase<Derived>& dense) { typename dense_traits<Derived>::T det = 0; // Laplace expansion for (std::size_t i = 0; i < N; ++i) det += ((i & 1) ? -1 : 1)*dense.coeff(0,i)*det_helper::run(dense.sub(0,i)); return det; } }; template <typename Derived> struct det_helper<Derived, 2> { static inline typename dense_traits<Derived>::T run(const Densebase<Derived>& dense) { return dense.coeff(0,0)*dense.coeff(1,1) - dense.coeff(0,1)*dense.coeff(1,0); } }; 

并这样调用：

T determinant() const
{
  return det_helper<Derived, dense_traits<Derived>::M>::run(derived());
}

• 第二个问题是当我在<< stream operator上实现重载时，该重载在Matrix上工作正常，但在stream << matrix.sub(0,0);上崩溃stream << matrix.sub(0,0); 甚至没有进入功能。

这是在PASTEBIN上上传的完整代码

并按要求输出错误：

main.cpp: In instantiation of 'static typename dense_traits<T>::T det_helper<Derived, N>::run(const Densebase<Derived>&) [with Derived = Matrix<float, 3ul, 3ul>; long unsigned int N = 3ul; typename dense_traits<T>::T = float]':
main.cpp:68:62:   required from 'typename Densebase<Derived>::base::T Densebase<Derived>::determinant() const [with Derived = Matrix<float, 3ul, 3ul>; typename Densebase<Derived>::base::T = float]'
main.cpp:159:30:   required from here
main.cpp:33:65: error: no matching function for call to 'det_helper<Matrix<float, 3ul, 3ul>, 3ul>::run(Densebase<Matrix<float, 3ul, 3ul> >::ConstSubmat)'
       det += ((i & 1) ? -1 : 1)*dense.coeff(0,i)*det_helper::run(dense.sub(0,i));
                                                                 ^
main.cpp:28:51: note: candidate: static typename dense_traits<T>::T det_helper<Derived, N>::run(const Densebase<Derived>&) [with Derived = Matrix<float, 3ul, 3ul>; long unsigned int N = 3ul; typename dense_traits<T>::T = float]
   static inline typename dense_traits<Derived>::T run(const Densebase<Derived>& dense)
                                                   ^
main.cpp:28:51: note:   no known conversion for argument 1 from 'Densebase<Matrix<float, 3ul, 3ul> >::ConstSubmat {aka const Subview<const Matrix<float, 3ul, 3ul>, 2ul, 2ul>}' to 'const Densebase<Matrix<float, 3ul, 3ul> >&'

Answer 1

我不知道您的编译错误是什么，但是Mathbase::coeff和Densebase::sub重载具有相同的签名（名称和参数类型），无法编译。 您不能仅通过函数的返回类型来重载它，这看起来就像您正在尝试做的那样。