将相同的参数转发给 C++ 中的可变元组构造函数

Question

TL；低于 DR

我正在尝试使用用于多核处理的包编写一些 C++ 代码。该包有一个很好的发件人 class，我用它在线程之间发送消息。它看起来像这样：

// structs provided by the package
struct GlobalVarsContainer {
    // some global vars
};

template<typename ...Ts>
struct sender {
    GlobalVarsContainer* my_container;
    sender (GlobalVarsContainer& c) : my_container(&c);
    void send(Ts... args) {
        // send the messages of types Ts
    };
};

这个class的构造函数实际上有点复杂，所以我希望只调用一次构造函数。我想为我需要的所有发件人使用容器类型，所以像这样：

typedef std::tuple<
    sender<uint32_t,uint32_t>,
    sender<uint32_t,double>,
    sender<uint32_t,double>
> sender_container_1_t;
typedef std::tuple<
    sender<uint32_t,double>,
    sender<uint32_t,double>,
    sender<uint32_t,double>
> sender_container_2_t;

现在我可以构造一个 sender_container_1_t 并通过引用传递给我的所有函数，欢呼！但这很难看，你必须像这样实例化容器：

GlobalVarsContainer gvc1, gvc2;
sender_container_1_t c1(gvc1, gvc1, gvc1);
sender_container_2_t c2(gvc2, gvc2, gvc2);

我想要一些形式

GlobalVarsContainer gvc1, gvc2;
// Should have a tuple of senders of type sender<uint32_t,uint32_t>, sender<uint32_t,double> and sender<uint32_t,double>, all constructed with a reference to gvc as an argument.
sender_container<uint32_t,double,double> c1(gvc1);
// Should have a tuple of senders of type sender<uint32_t,double>, sender<uint32_t, double> and sender<uint32_t, double>, all constructed with a reference to gvc as an argument.
sender_container<double,double,double> c2(gvc2);

所以我想到了使用可变容器结构，像这样：

// My sender container
template<typename... Ts>
struct SenderContainer {
    std::tuple<sender<uint32_t, Ts>...> my_senders;
    // What to do for the constructor ?
};

但我不知道该怎么做才能 SenderContainer<uint32_t,double,double> my_container(gvc1) 将 gvc1 转发给所有发件人的构造函数。有人有提示吗？

长话短说：如果我有可变结构模板

template<typename ...Ts>
struct Bar {
    Bar(Foo f){};
}

我可以做一个这样的容器吗

template<typename ...Ts>
struct BarContainer{
    std::tuple<Bar<int,Ts>...> bars
}

这样我就可以调用 BarContainer<int, double, long> newBarContainer(f) 并将 Foo f 转发给 BarContainer::bars 的所有构造函数。

Answer 1

您可以创建虚拟结构以允许您的 pack expansion:

template <typename> struct tag{};

template<typename ...Ts>
struct Bar {
    Bar(Foo f){};
};

template<typename ...Ts>
struct BarContainer{
    BarContainer(Foo f) : bars{(tag<Ts>{}, f)...} {}

    std::tuple<Bar<int,Ts>...> bars;
}

Demo

将相同的参数转发给 C++ 中的可变元组构造函数

Forward the same argument to variadic tuple constructor in C++

c++

templates

constructor

variadic-templates

stdtuple