提取函数参数类型作为参数包

Question

这是 "unpacking" a tuple to call a matching function pointer 的后续问题，它询问如何以通用方式将 std::tuple 中的值作为参数提供给函数。给出的解决方案如下：

template<int ...>
struct seq { };

template<int N, int ...S>
struct gens : gens<N-1, N-1, S...> { };

template<int ...S>
struct gens<0, S...>
{
   typedef seq<S...> type;
};

double foo(int x, float y, double z)
{
   return x + y + z;
}

template <typename... Args>
struct save_it_for_later
{
   std::tuple<Args...> params;
   double (*func)(Args...);

   double delayed_dispatch()
   {
    return callFunc(typename gens<sizeof...(Args)>::type());
   }

   template<int ...S>
   double callFunc(seq<S...>)
   {
    return func(std::get<S>(params) ...);
   }
};

int main(void)
{
   std::tuple<int, float, double> t = std::make_tuple(1, 1.2, 5);
   save_it_for_later<int,float, double> saved = {t, foo};
   std::cout << saved.delayed_dispatch() << std::endl;
}

我的问题是是否有办法制作 save_it_for_later 的替代版本，它只需要 foo 作为模板参数，这样我们就不必提供 foo ' s 参数类型作为模板参数（或将其 return 类型烘焙为 save_it_for_later）。像

int main(void) {
   ...
   save_it_for_later2<foo> saved = {t};
   ...
}

我同样可以使用某种宏包装 foo 来提取所需的类型：

int main(void) {
   ...
   save_it_for_later<MACRO_USING_DECLTYPE_OR_SOMESUCH(foo)> saved = {t};
   ...
}

这个问题似乎与原始问题正交，足以保证自己的票。

Answer 1

我只是羞怯地发现我去年问过一个类似的问题 (Unpacking arguments of a functional parameter to a C++ template class)，这里也给出了答案：

#include <functional>
#include <iostream>
#include <tuple>

template<int ...>
struct seq { };

template<int N, int ...S>
struct gens : gens<N-1, N-1, S...> { };

template<int ...S>
struct gens<0, S...>
{
   typedef seq<S...> type;
};

double foo(int x, float y, double z)
{
   return x + y + z;
}

template<typename T>
struct save_it_for_later;

template <typename Result, typename... Args>
struct save_it_for_later<Result(Args...)>
{
   std::tuple<Args...> params;
   Result (*func)(Args...);

   Result delayed_dispatch()
   {
    return callFunc(typename gens<sizeof...(Args)>::type());
   }

   template<int ...S>
   Result callFunc(seq<S...>)
   {
    return func(std::get<S>(params) ...);
   }
};

int main(void)
{
   std::tuple<int, float, double> t = std::make_tuple(1, 1.2, 5);
   save_it_for_later<decltype(foo)> saved = {t, foo};
   std::cout << saved.delayed_dispatch() << std::endl;
}

Answer 2

#include <tuple>
#include <utility>

template <typename> struct save_it_for_later_t;
template <typename Result, typename... Args>
struct save_it_for_later_t<Result (*)(Args...)> {
    std::tuple<Args...>   params;
    Result              (*fun)(Args...);
    template <typename... Params>
    save_it_for_later_t(Result (*fun)(Args...), Params&&... params)
        : params(std::forward<Params>(params)...)
        , fun(fun) {
    }
    // ... 
};
template <typename Result, typename... Args, typename... Params>
save_it_for_later_t<Result(*)(Args...)>
save_it_for_later(Result (*fun)(Args...), Params&&... params) {
    return save_it_for_later_t<Result(*)(Args...)>(fun, std::forward<Params>(params)...);
}

double foo(float, float, double);
int main() {
    auto saved = save_it_for_later(foo, 1.2f, 3.4f, 5.6);
    // ...
}

提取函数参数类型作为参数包

Extracting function argument types as a parameter pack

c++

function-pointers

variadic-templates

c++11