Question

我试图修改我的线性代数模块以避免虚拟vtable事情。试图使用CRTP和表达式模板。我选择了一些基本的东西来测试整个事情，但我无法让它发挥作用。

我有4节课，说：基础CRTP类，这里是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，我在Densebase中实现了transpose，determinant等函数：

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，它声明了两种类型，它们对父矩阵进行引用（co-matrix）然后我有Matrix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，有两件事情不起作用：

我实现了一个使用拉普拉斯扩展的行列式助手（计算共矩阵行列式并对它们求和，遗憾的是它无法编译

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);
   }
 };

并且这样称呼：

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

第二个问题是我在<< stream operator上实现了一个过载，它在Matrix上工作正常但在stream << matrix.sub(0,0);上崩溃，甚至没有输入该函数。

以下是PASTEBIN

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

我不知道你的编译错误是什么，但Mathbase::coeff和Densebase::sub的重载具有相同的签名（名称和参数类型），这些签名将无法编译。你不能仅仅通过它的返回类型来重载函数，它看起来就像你想要做的那样。

CRTP和表达式模板线性代数

1 个答案: