使用C ++元編程提取C函數的參數（“實用C ++元編程”中的示例）

Question

以下是“實用C ++元編程”（第16/17頁）中的示例：

#include <tuple>
#include <typeinfo>

template <typename F>
struct make_tuple_of_params;

template <typename Ret, typename... Args>
struct make_tuple_of_params<Ret (Args...)>
{
   using type = std::tuple<Args...>;
};

template <typename F>
using make_tuple_of_params_t = typename make_tuple_of_params<F>::type;

template<typename F>
void some_magic_function(F f)
{
   // if F is in the form void(double*, double*)
   // make_tuple_of_params is std::tuple<double*, double*>
   make_tuple_of_params_t<F> params;

   // ...
}

void Foo(double* x, double* y) { }

int main()
{
   some_magic_function(Foo);
}

它無法編譯：

$ clang++ -std=c++14 MakeTuple.cpp
MakeTuple.cpp:14:5: error: implicit instantiation of undefined template 'make_tuple_of_params<void (*)(double *, double*)>'

這是因為未定義make_tuple_of_params的非專用版本（上述代碼的第4行和第5行）嗎？

Answer 1

實際上，您需要進行不同的重載，具體取決於您是從指針還是從簽名模板參數中提取它，請參見下文。

template <typename F>
struct make_tuple_of_params;

template <typename Ret, typename... Args>
struct make_tuple_of_params<Ret (*)(Args...)> {
  using type = std::tuple<Args...>;
};

template <typename Ret, typename... Args>
struct make_tuple_of_params<Ret(Args...)> {
  using type = std::tuple<Args...>;
};

template <typename F>
using make_tuple_of_params_t = typename make_tuple_of_params<F>::type;

template <typename F>
bool some_magic_function(F f) {
  // if F is in the form void(double*, double*)
  // make_tuple_of_params is std::tuple<double*, double*>
  return std::is_same<std::tuple<double*, double*>, make_tuple_of_params_t<F>>::value;
}

void Foo(double* x, double* y) {}

int main() {
  cerr << some_magic_function(Foo) << endl;
  cerr
    << std::is_same<std::tuple<int, int>, make_tuple_of_params_t<void(int, int)>>::value
    << endl;
  // The latter one might be handy in some of template metaprogramming constructs
  return 0;
}

抱歉，沒有看過書頁，所以不知道授權在這里是什么意思。