如何从std :: tuple的类型获取元素 [英] How to get element from std::tuple by type

问题描述

我有一组类A，B，C和我想要从类型的通用代码访问实例fe

  template< typename T> 
 newObject（）
 {
 return m_storage-> getNew（）; 
} 
  

其中m_storage是A或B或C的实例， / p>

所以我想出了std :: tuple，但是有一个问题，因为我不能从类型tuple获取元素。

  std :: tuple< A，B，C> m_tpl; 
 template< typename T> 
 newObject（）
 {
 return m_tpl.get< T>（）。getNew（）; 
} 
  

有办法吗？

谢谢。

PS：
我不想为每个类型编写newObject的专业化： - ）

解决方案

这篇文章是C ++ 14关于从类型获得值的草案。

但在C之前++ 14会来，你可以写如下：

 命名空间详细信息
 {
 
 template< class T，std :: size_t N，class ... Args> 
 struct get_number_of_element_from_tuple_by_type_impl 
 {
 static constexpr auto value = N; 
}; 
 
 template< class T，std :: size_t N，class ... Args> 
 struct get_number_of_element_from_tuple_by_type_impl< T，N，T，Args ...> 
 {
 static constexpr auto value = N; 
}; 
 
 template< class T，std :: size_t N，class U，class ... Args> 
 struct get_number_of_element_from_tuple_by_type_impl< T，N，U，Args> 
 {
 static constexpr auto value = get_number_of_element_from_tuple_by_type_impl< T，N + 1，Args ...> :: value; 
}; 
 
} //命名空间细节
 
模板< class T，class ... Args> 
 T get_element_by_type（const std :: tuple< Args ...>& t）
 {
 return std :: get< detail :: get_number_of_element_from_tuple_by_type_impl< T，0，Args .. 。> :: value>（t）; 
} 
 
 int main（）
 {
 int a = 42; 
 
 auto t = std :: make_tuple（3.14，Hey！，std :: ref（a））; 
 
 get_element_by_type< int&>（t）= 43; 
 
 std :: cout<< a<< std :: endl; 
 
 // get_element_by_type< char>（t）; // tuple_element index out of range 
 
 return 0; 
} 
  

I have a set of classes A, B, C and I want to have access instances of them from generic code by type, f.e

template<typename T>
newObject()
{
    return m_storage->getNew();
}

where m_storage is instance of A or B or C, depends on T.

So I came up with std::tuple, but there is the problem because I can't get element from tuple by type.

std::tuple<A,B,C> m_tpl;
template<typename T>
newObject()
{
    return m_tpl.get<T>().getNew();
}

Is there any way to do it?Is this possible?

Thanks.

PS: I don't want to write the specialisation of newObject for each type.:-)

解决方案

This is a draft from C++14 about getting value from tuple by type.

But before C++14 will come, you could write something like below:

namespace detail
{

template <class T, std::size_t N, class... Args>
struct get_number_of_element_from_tuple_by_type_impl
{
    static constexpr auto value = N;
};

template <class T, std::size_t N, class... Args>
struct get_number_of_element_from_tuple_by_type_impl<T, N, T, Args...>
{
    static constexpr auto value = N;
};

template <class T, std::size_t N, class U, class... Args>
struct get_number_of_element_from_tuple_by_type_impl<T, N, U, Args...>
{
    static constexpr auto value = get_number_of_element_from_tuple_by_type_impl<T, N + 1, Args...>::value;
};

} // namespace detail

template <class T, class... Args>
T get_element_by_type(const std::tuple<Args...>& t)
{
    return std::get<detail::get_number_of_element_from_tuple_by_type_impl<T, 0, Args...>::value>(t);
}

int main()
{
    int a = 42;

    auto t = std::make_tuple(3.14, "Hey!", std::ref(a));

    get_element_by_type<int&>(t) = 43;

    std::cout << a << std::endl;

    // get_element_by_type<char>(t); // tuple_element index out of range

    return 0;
}

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