Question

我想生成＆＃39;函数指针的跳转表。指向的函数有两种类型的模板。应该为两个类型列表中的每个可能的对实例化一个不同的函数。理想情况下，我们可以有类似的东西：

#include <tuple>

template <typename X, typename Y>
void foo()
{}

template <typename... Xs, typename... Ys>
void bar(const std::tuple<Xs...>&, const std::tuple<Ys...>&)
{
  using fun_ptr_type = void (*) (void);
  static constexpr fun_ptr_type jump_table[sizeof...(Xs) * sizeof...(Ys)]
    = {&foo<Xs, Ys>...};
}

int main ()
{
  using tuple0 = std::tuple<int, char, double>;
  using tuple1 = std::tuple<float, unsigned long>;

  bar(tuple0{}, tuple1{});
}

正如预期的那样，当元组具有不同的长度时它会失败：

foo.cc:15:20: error: pack expansion contains parameter packs 'Xs' and 'Ys' that have different lengths (3 vs. 2)
    = {&foo<Xs, Ys>...};
            ~~  ~~ ^
foo.cc:23:3: note: in instantiation of function template specialization 'bar<int, char, double, float, unsigned long>' requested here
  bar(tuple0{}, tuple1{});
  ^
1 error generated.

为了实现这种功能，我已经尝试并成功使用indirection（第一个跳转表，其中包含指向另一个跳转表的函数的指针），但我发现它很笨拙。

所以，我的问题是：是否有解决方法？

Answer 1

您的示例代码是错误的，即使它编译的情况（即sizeof ...（Xs）== sizeof ...（Ys））。比方说，你有N元元组，然后jump_table有N * N个元素，但只有前N个元素用函数ptrs初始化。

首先，您需要内部加入2个类型列表：

template<class A, class B>
struct P;

template<class... Ts>
struct L {};

template<class T, class... Ts>
using mul = L<P<T, Ts>...>;

template<class...>
struct cat;

template<class T>
struct cat<T>
{
    using type = T;
};

template<class... As, class... Bs>
struct cat<L<As...>, L<Bs...>>
{
    using type = L<As..., Bs...>;
};

template<class A, class B, class... Ts>
struct cat<A, B, Ts...>
{
    using type = typename cat<typename cat<A, B>::type, Ts...>::type;
};

template<class A, class B>
struct join;

template<class... As, class... Bs>
struct join<L<As...>, L<Bs...>>
{
    using type = typename cat<mul<As, Bs...>...>::type;
};

例如，

join<L<int[1], int[2]>, L<float[1], float[2], float[3]>>::type

给你

L<P<int[1], float[1]>, P<int[1], float[2]>, P<int[1], float[3]>, P<int[2], float[1]>, P<int[2], float[2]>, P<int[2], float[3]>

回到你的例子：

template <typename X, typename Y>
void foo()
{}

template<class T, std::size_t N>
struct jump_table
{
    template<class... As, class... Bs>
    constexpr jump_table(L<P<As, Bs>...>)
      : table{&foo<As, Bs>...}
    {}

    T table[N];
};

template <typename... Xs, typename... Ys>
void bar(const std::tuple<Xs...>&, const std::tuple<Ys...>&)
{
  using fun_ptr_type = void (*) (void);
  static constexpr jump_table<fun_ptr_type, sizeof...(Xs) * sizeof...(Ys)> table
    = {typename join<L<Xs...>, L<Ys...>>::type()};
}

int main ()
{
  using tuple0 = std::tuple<int, char, double>;
  using tuple1 = std::tuple<float, unsigned long>;

  bar(tuple0{}, tuple1{});
}

这应该符合您的预期。

Answer 2

这里的其他答案对于手头的问题来说似乎太复杂了。以下是我的表现：

#include <array>
#include <tuple>

template <typename X, typename Y> void foo() {}

using fun_ptr_type = void (*) (void);

// Build one level of the table.
template <typename X, typename ...Ys>
constexpr std::array<fun_ptr_type, sizeof...(Ys)>
  jump_table_inner = {{&foo<X, Ys>...}};

// Type doesn't matter, we're just declaring a primary template that we're
// about to partially specialize.
template <typename X, typename Y> void *jump_table;

// Build the complete table.
template <typename ...Xs, typename ...Ys>
constexpr std::array<std::array<fun_ptr_type, sizeof...(Ys)>, sizeof...(Xs)>
  jump_table<std::tuple<Xs...>, std::tuple<Ys...>> = {jump_table_inner<Xs, Ys...>...};

int main () {
  using tuple0 = std::tuple<int, char, double>;
  using tuple1 = std::tuple<float, unsigned long>;

  // Call function for (int, float).
  jump_table<tuple0, tuple1>[0][0]();
}

这是Clang 3.5在C ++ 14模式下接受的。

Answer 3

我正常的产品扩展解决方案context( f<Xs, Ys>... ) /* not what we want */是将其重写为context2( g<Xs, Ys...>... )。这意味着g负责针对某些Ys展开X，并且最终展开针对所有g执行Xs。这种重写的结果是我们引入了额外的嵌套，因此引入了不同的上下文。

在我们的例子中，我们将有一个函数指针数组，而不是一个平面的函数指针数组。 与您尝试的解决方案不同，虽然这些解决方案实际上是我们关心的&foo<X, Y>函数指针 - 并且展平是直截了当的。

#include <cassert> #include <utility> #include <array> template<typename X, typename Y> void foo() {} using foo_type = void(*)(); template<typename... T> struct list { static constexpr auto size = sizeof...(T); }; template<typename X, typename Y, typename Indices = std::make_index_sequence<X::size * Y::size>> struct dispatch; template< template<typename...> class XList, typename... Xs , template<typename...> class YList, typename... Ys , std::size_t... Indices > struct dispatch<XList<Xs...>, YList<Ys...>, std::index_sequence<Indices...>> { private: static constexpr auto stride = sizeof...(Ys); using inner_type = std::array<foo_type, stride>; using multi_type = inner_type[sizeof...(Xs)]; template<typename X, typename... Yss> static constexpr inner_type inner() { return {{ &foo<X, Yss>... }}; } static constexpr multi_type multi_value = { inner<Xs, Ys...>()... }; public: static constexpr auto size = sizeof...(Xs) * sizeof...(Ys); static constexpr foo_type value[size] = { multi_value[Indices / stride][Indices % stride]... }; }; template< template<typename...> class XList, typename... Xs , template<typename...> class YList, typename... Ys , std::size_t... Indices > constexpr foo_type dispatch<XList<Xs...>, YList<Ys...>, std::index_sequence<Indices...>>::value[size]; int main() { using dispatch_t = dispatch< list<int, char, double>, list<float, unsigned long> >; constexpr auto&& table = dispatch_t::value; static_assert( dispatch_t::size == 6, "" ); static_assert( table[0] == &foo<int, float>, "" ); static_assert( table[1] == &foo<int, unsigned long>, "" ); static_assert( table[2] == &foo<char, float>, "" ); static_assert( table[3] == &foo<char, unsigned long>, "" ); static_assert( table[4] == &foo<double, float>, "" ); static_assert( table[5] == &foo<double, unsigned long>, "" ); }

Coliru demo

Answer 4

你所拥有的实际上更像是两个列表（<X1,Y1>，<X2,Y2>，......）的“zip”，当列表的长度不同时，它们不起作用。

要计算两者的“产品”，我认为你必须使用辅助类才能使它工作。看到像你这样的其他问题：How to create the Cartesian product of a type list?

展开具有不同长度的参数包

4 个答案: