Implicit type conversion rules in C++ operators

Lists the implicit type conversion for C++.

So the question is, is there a type trait that can do that table? something akin to

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

(Not my question, but this doesn't compile:

/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)

)

有帮助吗?

解决方案

How about decltype(std::declval<T>() + std::declval<J>())?

许可以下: CC-BY-SA归因
不隶属于 StackOverflow
scroll top