c++ - 使用 C++ 元编程提取 C 函数的参数(来自 "Practical C++ Metaprogramming"的示例)

Question

以下是“实用 C++ 元编程”(第 16/17 页)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);
}

编译失败:

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

这是不是因为make_tuple_of_params的非特化版本(上面代码的第4行和第5行)没有定义？

Answer 1

您实际上需要不同的重载，具体取决于您是从指针还是从签名模板参数中提取它，请参见下文。

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;
}

不好意思，没看过书页，不知道作者的意思。