使用关键字using在C ++中声明模板化函数指针的语法

Question

考虑

template<typename T>
struct auxiliary
{
  typedef std::unique_ptr<T> (*creator)(std::vector<double>const&);
};
template<typename T>
using creator = typename auxiliary<T>::creator;

我想知道如何宣布没有auxiliary creator ，即

template<typename T>
using creator = ???    

Answer 1

template<typename T>
using creator = std::unique_ptr<T> (*)(std::vector<double> const&);