用 constexpr if 定义和声明一个 const std::array

Question

我正在尝试实现 Gauss-Legendre 求积，我想要一个模板函数 它将点数作为模板参数。 现在我有这个：

template<int number_of_quadrature_points>
double gaussian_quadrature_integral_core(double (*f)(double), double from, double to){
    double scaling_factor = (to-from)/2;   
    double average_factor = (from+to)/2;
    std::array<double, number_of_quadrature_points> factors;
    std::array<double, number_of_quadrature_points> points;
    if constexpr(number_of_quadrature_points == 2){
        factors = {1, 1};
        points = {-1.0/sqrt(3), 1.0/sqrt(3)};
    }
    if constexpr(number_of_quadrature_points == 3){
        factors = {5.0/9.0, 8.0/9.0, 5.0/9.0};
        points = {-sqrt(3.0/5.0), 0, sqrt(3.0/5.0)};
    }
    
    double sum = 0;

    for(int i = 0; i < number_of_quadrature_points; i++){
        sum += factors.at(i)*((*f)(scaling_factor*points.at(i)+average_factor));
    }

    sum *= scaling_factor;
    return sum;
}

如你所见，当模板参数改变时，不仅数组大小改变，内容也改变，但对于给定大小，内容是众所周知的。出于这个原因，我认为如果 std::arrays 是 const static 会更好，因为该函数会被多次调用。

现在我只设法使用 if constexpr 来声明数组，但是我怎样才能使用它来定义和声明数组，以便它在 if constexpr 范围之外可见并且数组只定义一次？

Answer 1

添加两个辅助函数就足够了（如果您使用的是 C++20）：

template<unsigned N>
constexpr auto init_factors() {
    std::array<double, N> rv;
    if constexpr(N == 2){
        rv = {1., 1.};
    } else {
        rv = {5.0/9.0, 8.0/9.0, 5.0/9.0};
    }
    return rv;
}

template<unsigned N>
constexpr auto init_points() {
    std::array<double, N> rv;
    if constexpr(N == 2){
        rv = {-1.0/std::sqrt(3.), 1.0/std::sqrt(3.)};
    } else {
        rv = {-std::sqrt(3.0/5.0), 0, std::sqrt(3.0/5.0)};
    }
    return rv;
}

template<unsigned number_of_quadrature_points>
double gaussian_quadrature_integral_core(double (*f)(double), double from,
                                                              double to)
{
    static constexpr auto factors = init_factors<number_of_quadrature_points>();
    static constexpr auto points = init_points<number_of_quadrature_points>();
[...]

为防止使用错误的点数，您可以添加static_assert

template<unsigned number_of_quadrature_points>
double
gaussian_quadrature_integral_core(double (*f)(double), double from,
                                                       double to)
{
    static_assert(number_of_quadrature_points==2||number_of_quadrature_points==3);

...如果您想稍后进行专业化，则使用 SFINAE 阻止匹配：

#include <type_traits>

template<unsigned number_of_quadrature_points>
std::enable_if_t<number_of_quadrature_points==2||number_of_quadrature_points==3,
                 double>
gaussian_quadrature_integral_core(double (*f)(double), double from,
                                                       double to)
{

Answer 2

你可能有模板变量：

template <std::size_t N>
static constexpr std::array<double, N> factors;

template <std::size_t N>
static constexpr std::array<double, N> points;

template <>
constexpr std::array<double, 2> factors<2>{{1, 1}};
template <>
constexpr std::array<double, 2> points<2>{{-1.0 / sqrt(3), 1.0 / sqrt(3)}};

template <>
constexpr std::array<double, 3> factors<3>{{5.0 / 9.0, 8.0 / 9.0, 5.0 / 9.0}};
template <>
constexpr std::array<double, 3> points<3>{{-sqrt(3.0 / 5.0), 0, sqrt(3.0 / 5.0)}};

然后

template<int number_of_quadrature_points>
double gaussian_quadrature_integral_core(double (*f)(double), double from, double to)
{
    const double scaling_factor = (to - from) / 2;   
    const double average_factor = (from + to) / 2;
    double sum = 0;

    for(int i = 0; i < number_of_quadrature_points; i++){
        sum += factors<number_of_quadrature_points>[i]
           * ((*f)(scaling_factor * points<number_of_quadrature_points>[i] + average_factor));
    }

    sum *= scaling_factor;
    return sum;
}

Demo

请注意，如果没有 constexpr sqrt（std:: 没有），则必须将 constexpr 替换为 const。

Answer 3

您可以在本主题中使用类似的内容： is there a way to put condition on constant value parameter in C++ template specialization?

因此，我们使用std::enable_if 和 SFINAE 创建了两个模板特化。我们通过模板参数number_of_quadrature_points来区分它们。这样我们就有了不需要多次定义和实例化的全局参数。此代码使用 c++17 编译。

另外我建议使用带有std::function<> 的现代方法，而不是指向函数的指针。

#include <array>
#include <cmath>
#include <iostream>
#include <functional>

template<int number_of_quadrature_points, typename E=void>
struct gaussian_quadrature_params
{

};

template<int number_of_quadrature_points>
struct gaussian_quadrature_params<number_of_quadrature_points, std::enable_if_t<(number_of_quadrature_points==2)> >
{
    constexpr static const std::array<double, number_of_quadrature_points> factors = {1, 1};
    constexpr static const std::array<double, number_of_quadrature_points> points = {-1.0/sqrt(3), 1.0/sqrt(3)};
};

template<int number_of_quadrature_points>
struct gaussian_quadrature_params<number_of_quadrature_points, std::enable_if_t<(number_of_quadrature_points==3)> >
{
    constexpr static const std::array<double, number_of_quadrature_points> factors = {5.0/9.0, 8.0/9.0, 5.0/9.0};
    constexpr static const std::array<double, number_of_quadrature_points> points = {-sqrt(3.0/5.0), 0, sqrt(3.0/5.0)};
};


double f(double x)
{
    return x;
}


template<int number_of_quadrature_points>
double gaussian_quadrature_integral_core(std::function<double(double)> f, double from, double to){
    double scaling_factor = (to-from)/2;   
    double average_factor = (from+to)/2;
    
    double sum = 0;

    for(int i = 0; i < number_of_quadrature_points; i++){
        sum += gaussian_quadrature_params<number_of_quadrature_points>::factors.at(i)*(f(scaling_factor*gaussian_quadrature_params<number_of_quadrature_points>::points.at(i)+average_factor));
    }

    sum *= scaling_factor;
    return sum;
}

int main()
{
   std::cout << gaussian_quadrature_integral_core<2>(f, -1.0, 1.0) << std::endl;
   std::cout << gaussian_quadrature_integral_core<3>(f, -1.0, 1.0) << std::endl;
}