继承CTAD构造函数

Question

我是从std::tuple派生的，但由于类模板参数演绎的问题，无法从初始化程序列表构造派生类。是否有更好的方法来构造这样一个类，而不仅仅是给它一个已经构造的元组first{ std::tuple{1, 1.0f, 1u} };。

template <typename T, typename... Types>
struct first : public std::tuple<T, Types...>
{
    //using std::tuple<T, Types...>::tuple;
    template<typename F>
        requires std::invocable<F, T>
    auto transform(F f)
    {
        auto result = *this;
        std::get<0>(result) = f(std::get<0>(result));
        return result;
    }
};

int main()
{
    //auto tuple = first{ 1, 1.0f, 1u };
    auto tuple = first{ std::tuple{1, 1.0f, 1u} };
    auto tuple2 = tuple.transform([](auto a) {return a + 3; });
}

Answer 1

继承的构造函数不是CTAD的一部分。

你可以复制std:元组CTAD

template <typename T, typename... Types>
first(T, Types...) -> first<T, Types...>;

// auto tuple = first{1, 1.0f, 1u}; // is ok now
// auto tuple = first{ std::tuple{1, 1.0f, 1u} }; // broken now

演示

Answer 2

我不能确切地告诉您为什么using std::tuple<T, Types...>::tuple;指令在这里不能工作，但是如果您只定义接受参数T, Types...和手动接受std::tuple<T, Types...>的构造函数，那么一切似乎都很好：

template <typename T, typename... Types>
struct first : public std::tuple<T, Types...>
{
    first(T&& t, Types&&... vals)
        : std::tuple<T, Types...>{t, std::forward<Types>(vals)...}
    {}

    first(std::tuple<T, Types...>&& t)
        : std::tuple<T, Types...>(t)
    {}

    template<typename F>
        requires std::invocable<F, T>
    auto transform(F f)
    {
        auto result = *this;
        std::get<0>(result) = f(std::get<0>(result));
        return result;
    }
};

int main()
{
    auto tuple1 = first{1, 1.0f, 1u};
    auto tuple2 = first{std::tuple{1, 1.0f, 1u}};
    auto tuple3 = tuple2.transform([](auto a) {return a + 3; });

    std::cout << std::get<0>(tuple1) << ", " << std::get<1>(tuple1) << ", " << std::get<2>(tuple1) << std::endl;
    std::cout << std::get<0>(tuple3) << ", " << std::get<1>(tuple3) << ", " << std::get<2>(tuple3) << std::endl;
}

产出如下：

1, 1, 1
4, 1, 1

这就是你想要的？