"Merge"PODs合二为一

Question

有没有办法在它们之间组合（或合并、聚合）PODs？一个直观的解决方案：

struct Base1 { int i; };
struct Base2 { char c; };

struct Derived : Base1, Base2 {};

// in a more generalized way
// template <class... Ts> struct Aggregate : Ts... {};

除了，我们失去了一些东西：

int main()
{
    Derived d {42, 'c'}; // OK
    auto [i, c] = d; // Error
    static_assert(std::is_standard_layout_v<Derived>); // Error
}

我知道合并后的基础类之间可能会出现一些歧义和冲突。但是将 PODs 合并为一个会非常好。我尝试达到的结果：

struct Expected { int i; char c; }; // I want Derived to behave exactly like this

我在反思的境界吗？我应该最终使用宏吗？

Answer 1

如果我们

don't care if the derived struct is a POD or not

，Boost.PFR 的任务非常简单 - 只需将您的 PODs 转换为元组并连接它们：

template<typename... Ts> using Merge = decltype(std::tuple_cat(
    boost::pfr::structure_to_tuple(std::declval<Ts>())...
));

一个简单的测试：

int main() {
    struct Base1 { int i, j; };
    struct Base2 { char c; };
    using Expected = Merge<Base1, Base2>;
    static_assert(std::is_same_v<Expected, std::tuple<int, int, char>>);
    Expected expected{42, 1337, 'c'};
    auto&& [i, j, c] = expected;
}

"Merge"PODs合二为一

"Merge" PODs into one

c++

inheritance

reflection

structured-bindings

standard-layout