问题描述
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?
Member1
和 Member2
没有默认/复制/移动构造函数。
我知道 std::apply
,正如 How do I expand a tuple into variadic template function's arguments? 中所建议的。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。
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
,向它传递两组可变参数以完美转发它们以初始化 member1
和 member2
。我认为将两个集合“分组”为元组可能是一个好主意,因为 std::map::emplace
就是这样做的。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)。
解决方法
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,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'});
}
,
cppreference.com has a nice example of a sample implementation of std::make_from_tuple
,然而,正如您发现的,由于缺少底层类的复制构造函数,您无法使用它。
但是,它的面包屑可以让您调整它以解决这些限制:
#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;
}
使用带有 -std=c++17
的 gcc 10 进行测试。