如何将元组的值转发给成员初始值设定项？

Question

我需要将元组的值转发给成员初始值设定项：

struct Struct {
  Member1 member1;
  Member2 member2;

  template<typename Tuple1, typename Tuple2>
    Struct( Tuple1&& tuple1, Tuple2&& tuple2 )
      : member1(tuple1...), member2(tuple2...)
    {}
};

上面的代码显然是无效的。怎么表达呢？

Member1 和 Member2 没有 default/copy/move 构造函数。

我知道 std::apply，如 How do I expand a tuple into variadic template function's arguments? 中所述。我也知道 std::make_from_tuple。但是我不知道如何在成员初始值设定项中使用这些中的任何一个。

任何 C++ 标准都可以（最好是 C++17，但 C++20 也可以）。

澄清一下，我的真正目标是创建一个 Struct，将两组可变参数传递给它以完美转发它们以初始化 member1 和 member2。我认为将两个集合“分组”成元组可能是个好主意，因为那是 std::map::emplace 所做的。其他方法也可以工作（例如，在两组可变参数之间传递一个特殊对象）。

Answer 1

cppreference.com has a nice example of a sample implementation of std::make_from_tuple，但是，正如您所发现的，由于缺少底层 class.[=14= 的复制构造函数，您无法使用它]

但是，它的面包屑导航可让您对其进行调整以解决这些限制：

#include <tuple>
#include <iostream>

struct Member1 {
    Member1(int a, int b)
    {
        std::cout << a << " "
              << b
              << std::endl;
    }

    Member1(const Member1 &)=delete;
    Member1(Member1 &&)=delete;
};

struct Member2 {
    Member2(const char *str)
    {
        std::cout << str << std::endl;
    }

    Member2(const Member2 &)=delete;
    Member2(Member2 &&)=delete;
};

// De-obfucation shortcut

template<typename Tuple>
using make_index_sequence_helper=std::make_index_sequence
    <std::tuple_size_v<std::remove_reference_t<Tuple>>>;

struct Struct {
    Member1 member1;
    Member2 member2;

    template<typename Tuple1, typename Tuple2>
    Struct( Tuple1&& tuple1,
        Tuple2&& tuple2 )
        : Struct{std::forward<Tuple1>(tuple1),
        make_index_sequence_helper<Tuple1>{},
        std::forward<Tuple2>(tuple2),
        make_index_sequence_helper<Tuple2>{}}
    {
    }

    template<typename Tuple1, std::size_t ...tuple1_args,
         typename Tuple2, std::size_t ...tuple2_args>
    Struct(Tuple1 && tuple1,
           std::index_sequence<tuple1_args...>,
           Tuple2 && tuple2,
           std::index_sequence<tuple2_args...>)
        : member1{std::get<tuple1_args>(tuple1)...},
          member2{std::get<tuple2_args>(tuple2)...}
    {
    }
};

int main()
{
    Struct s{ std::tuple<int, int>{2, 3},
        std::tuple<const char *>{"Hello world"}};

    return 0;
}

使用 gcc 10 和 -std=c++17 进行测试。

Answer 2

std::make_from_tuple确实是正确的选择：

#include <tuple>
struct Member1 {
    Member1(int x,float y, char z){}

    Member1(const Member1& other)=delete;
    Member1(Member1&& other)=delete;
};

struct Member2 {
    Member2(int x,float y, char z){}

    Member2(const Member2& other)=delete;
    Member2(Member2&& other)=delete;
};

struct Struct {
  Member1 member1;
  Member2 member2;

  template<typename Tuple1, typename Tuple2>
    Struct(Tuple1&& tuple1, Tuple2&& tuple2)
      : member1(std::make_from_tuple<Member1>(std::forward<Tuple1>(tuple1))),
       member2(std::make_from_tuple<Member2>(std::forward<Tuple2>(tuple2)))
    {}
};

int main(){
    Struct c(std::tuple{1,1.1,'c'},std::tuple{2,2.2,'x'});
}

Godbolt demo.

如何将元组的值转发给成员初始值设定项？

How do I forward the values of tuple to a member initializer?

c++

tuples

member-initialization

perfect-forwarding