如何将元组的值转发给成员初始值设定项? [英] How do I forward the values of tuple to a member initializer?

问题描述

我需要将元组的值转发给成员初始值设定项:

I need to forward the values of a tuple to a member initializer:

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?

The code above obviously isn't valid. How can I express it?

Member1 和 Member2 没有默认/复制/移动构造函数.

Member1 and Member2 have no default/copy/move constructor.

我了解 std::apply，正如 如何将元组扩展为可变参数模板函数的参数?.我也知道 std::make_from_tuple.但我不知道如何在成员初始值设定项中使用这些.

I know about std::apply, as suggested in How do I expand a tuple into variadic template function's arguments?. I also know about std::make_from_tuple. But I wouldn't know how to use any of these in a member initializer.

任何 C++ 标准都可以(最好是 C++17，但 C++20 也可以).

Any C++ standard is fine (preferably C++17, but C++20 would work as well).

澄清一下，我的真正目标是创建一个 Struct，向它传递两组可变参数以完美转发它们以初始化 member1 和 member2.我以为分组"将两组组合成元组可能是一个好主意，因为这就是 std::map::emplace 所做的.其他方法也可以(例如，在两组可变参数之间传递一个特殊对象).

To clarify, my real goal is to create a Struct, passing it two sets of variadic arguments to perfect-forward them to initialize member1 and member2. I thought that "grouping" the two sets into tuples could have been a good idea, since that's what std::map::emplace does. Other approaches would work as well (e.g. passing a special object between the two sets of variadic arguments).

推荐答案

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 演示.

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