单个函数中的多个参数包？

Question

我正在尝试创建一个带有两个对象参数包的函数。有两个模板化基类，我想将派生类的实例传递给此函数。考虑这个例子。

template <int N>
struct First {};

template <int N>
struct Second {};

// there are a few of these
struct FirstImpl : First<5> {};
struct SecondImpl : Second<7> {};

template <int... firstInts, int... secondInts>
void function(float f, First<firstInts> &... first, Second<secondInts> &... second) {
  // ...
}

我想做的就是function像这样打电话

FirstImpl firstImpl;
OtherFirstImpl otherFirstImpl;
SecondImpl secondImpl;
OtherSecondImpl otherSecondImpl;
function(9.5f, firstImpl, otherFirstImpl, secondImpl, otherSecondImpl);

但这个例子无法编译。编译器似乎试图将所有内容打包到第二个参数包中，但由于FirstImpl无法隐式转换而失败Second<N>。

我该如何解决这个问题？

Answer 1

让我们首先编写一个变量模板来确定类型是否派生自First：

template <int N>
constexpr std::true_type is_first(First<N> const &) { return {}; }
template <int N>
constexpr std::false_type is_first(Second<N> const &) { return {}; }

template <class T>
constexpr bool is_first_v = decltype( is_first(std::declval<T>()) )::value;

以及一个收集和 的Split索引的结构FirstSecond结构：

template <class, class, class, std::size_t I = 0> struct Split;

template <
    std::size_t... FirstInts,
    std::size_t... SecondInts,
    std::size_t N
>
struct Split<
    std::index_sequence<FirstInts...>,
    std::index_sequence<SecondInts...>,
    std::tuple<>,
    N
> {
    using firsts = std::index_sequence<FirstInts...>;
    using seconds = std::index_sequence<SecondInts...>;
};

template <
    std::size_t... FirstInts,
    std::size_t... SecondInts,
    std::size_t I, 
    typename T,
    typename... Tail
>
struct Split<
    std::index_sequence<FirstInts...>,
    std::index_sequence<SecondInts...>,
    std::tuple<T, Tail...>,
    I
> : std::conditional_t<
    is_first_v<T>,
    Split<std::index_sequence<FirstInts..., I>,
          std::index_sequence<SecondInts...>,
          std::tuple<Tail...>,
          I + 1
    >,
    Split<std::index_sequence<FirstInts...>,
          std::index_sequence<SecondInts..., I>,
          std::tuple<Tail...>,
          I + 1
    >
> {};

就像我在评论中告诉你的那样，value向Firstand添加成员Second（或从 继承std:integral_constant），这允许我们编写以下内容：

template <std::size_t... FirstIdx, std::size_t... SecondIdx, typename Tuple>
void function_impl(float f, std::index_sequence<FirstIdx...>, std::index_sequence<SecondIdx...>, Tuple const & tup) {
    ((std::cout << "firstInts: ") << ... << std::get<FirstIdx>(tup).value) << '\n';
    ((std::cout << "secondInts: ") << ... << std::get<SecondIdx>(tup).value) << '\n';
    // your implementation
}

template <class... Args>
void function(float f, Args&&... args)  {
    using split = Split<std::index_sequence<>,std::index_sequence<>, std::tuple<std::decay_t<Args>...>>;
    function_impl(f, typename split::firsts{}, typename split::seconds{}, std::forward_as_tuple(args...));
}

Demo