使用Variadic模板的任意维数组 [英] Arbitrary dimensional array using Variadic templates

问题描述

如何在C ++ 11中创建一个Array类，可以像

一样使用

  Array< int，2,3,4> a，b; 
 Array< char，3,4> d; 
 Array<短，2> e; 
  

并以

<$> p $ p> a [2] [1] [2] = 15;
d [1] [2] ='a';


我还需要重载operator

  T& operator [size_t i_1] [size_t i_2] ... [size_t i_D]; 
  

不存在。

解决方案

最简单的方法是通过嵌套 std :: array ：

 ＃include< array& 
 
 template< class T，size_t size，size_t ... sizes> 
 struct ArrayImpl {
 using type = std :: array< typename ArrayImpl< T，sizes ...> :: type，size> ;; 
}; 
 
 template< class T，size_t size> 
 struct ArrayImpl< T，size> {
 using type = std :: array< T，size> ;; 
}; 
 
 template< class T，size_t ... sizes> 
 using Array = typename ArrayImpl< T，sizes ...> :: type; 
 

在此解决方案中 Array< char，3，4> 与 std :: array< std :: array< char，4>，3> 相同。

这也显示了如何为多个维度实现 operator [] 。 运算符[] 需要返回为其定义 operator [] 的对象。在这种情况下，它引用一个较小维数的数组。

How can I create an Array class in C++11 which can be used like

Array < int, 2, 3, 4> a, b; 
Array < char, 3, 4> d; 
Array < short, 2> e;

and access it in a way like

a[2][1][2] = 15; 
d[1][2] ='a';

I also need to overload operator as

T &operator[size_t i_1][size_t i_2]...[size_t i_D]; 

which does not exist. How can I do this?

解决方案

The simplest way to do this is by nesting std::array:

#include<array>

template<class T, size_t size, size_t... sizes>
struct ArrayImpl {
    using type = std::array<typename ArrayImpl<T, sizes...>::type, size>;
};

template<class T, size_t size>
struct ArrayImpl<T, size> {
    using type = std::array<T, size>;
};

template<class T, size_t... sizes>
using Array = typename ArrayImpl<T, sizes...>::type;

In this solution Array<char, 3, 4> is the same as std::array<std::array<char, 4>, 3> - array consisting of arrays of smaller dimension.

This also shows how you can implement operator[] for many dimensions. operator[] of your object needs to return object for which operator[] is also defined. In this case it is reference to an array of smaller dimension.

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