通过constexpr或模板函数在编译时获取多维std :: array的大小 [英] Get the size of multi dimensional std::array at compile time via constexpr or template function

问题描述

我使用三维 std :: array ，因为在编译时已经知道大小。但是，我注意到size（）函数不是静态的，因此constexpr / template函数无法访问。

I use a three dimensional std::array, because the size is already known at compile. However, I noticed that the size() function is not static, and thus, inaccessible for constexpr/template functions.

我已经找到下面的演示示例，该示例估计一维 std :: array 的大小。但是，这不适用于两个或多个维度。有没有一种方法可以通过为 x，y，z，...编写带有附加模板参数 dim 的函数来返回其他维度。 code>维？

I already found the demo example below, which estimates the size of a one dimensional std::array. However, this does not work for two or more dimensions. Is there a way to return the other dimensions by writing a function with an additional template parameter dim for the x, y, z, .. dimension?

// Example program
#include <iostream>
#include <string>
#include <array>

// typedefs for certain container classes
template<class T, size_t x>
using array1D = std::array<T, x>;
template<class T, size_t x, size_t y>
using array2D = std::array<std::array<T, y>, x>;
template<class T, size_t x, size_t y, size_t z>
using array3D = std::array<std::array<std::array<T, z>, y>, x>;


template<class T, std::size_t N>
auto array_size_helper(const array1D<T, N>&) -> std::integral_constant<std::size_t, N>;

template<class Array>
using array_size = decltype(array_size_helper(std::declval<const Array&>()));

template<class Array>
constexpr auto static_size() -> decltype(array_size<Array>::value) {
  return array_size<Array>::value;
}
template<class Array>
constexpr auto static_size(Array const&) -> decltype(static_size<Array>()) {
  return static_size<Array>();
}

int main()
{
    std::cout << static_size<array3D<float, 3, 4, 5>>();
}

推荐答案

对于一维情况，您还可以使用为 std :: array 定义的 std :: tuple_size ：

For the one dimensional case, you could use std::tuple_size which is defined for std::array as well:

int main()
{
    std::cout << std::tuple_size<array3D<float, 3, 4, 5>>();
}

关于您的实际问题。如果我理解的正确，那么您希望在size函数上有一个附加参数，您可以通过该参数选择要返回尺寸的尺寸，对吗？

Regarding your actual problem. If I understand you right you want to have an additional parameter on your size function with which you can select the dimension for which the size should be returned, right?

这可以使用递归可以轻松完成。这是一个工作示例：

This can be done easily by using recursion. Here is a working example:

// Example program
#include <iostream>
#include <string>
#include <array>

// typedefs for certain container classes
template<class T, size_t x>
using array1D = std::array<T, x>;

template<class T, size_t x, size_t y>
using array2D = std::array<std::array<T, y>, x>;

template<class T, size_t x, size_t y, size_t z>
using array3D = std::array<std::array<std::array<T, z>, y>, x>;


template <size_t dim, typename Array>
struct size_of_dim;

// specialization for std array and first dimension
template <typename T, size_t N>
struct size_of_dim<0, std::array<T,N>> : std::integral_constant<size_t, N> {};

// specialization for std array and dimension > 0 → recurse down in dim
template <size_t dim, typename InnerArray, size_t N>
struct size_of_dim<dim, std::array<InnerArray,N>> : size_of_dim<dim-1,InnerArray> {};



int main()
{
    std::cout << size_of_dim<0,array3D<float, 3, 4, 5>>() << std::endl;
    std::cout << size_of_dim<1,array3D<float, 3, 4, 5>>() << std::endl;
    std::cout << size_of_dim<2,array3D<float, 3, 4, 5>>() << std::endl;
}

演示

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