c++ - 键入特征以从运算符(operator)处获取提升结果

Question

Implicit type conversion rules in C++ operators

列出 C++ 的隐式类型转换。

那么问题来了，是否有类型特征可以做那个表呢？类似于

template <typename T, typename J>
struct promotion_type
{
  typedef decltype(operator+(const T&,const J&)) type;
};

(这不是我的问题，但这无法编译:

/home/user/source/testdir/main.cpp:97:51: error: there are no arguments to 'operator+' that depend on a template parameter, so a declaration of 'operator+' must be available [-fpermissive]
   typedef decltype(operator+(const T&,const J&)) type;
                                               ^
/home/user/source/testdir/main.cpp:97:51: note: (if you use '-fpermissive', G++ will accept your code, but allowing the use of an undeclared name is deprecated)

)

Answer 1

decltype(std::declval<T>() + std::declval<J>())怎么样？ ？