“_HAS_CXX17”marco是否可用于自定义项目标题以启用C ++ 17语言集功能？

Question

I want to create headers which use 'optional' from standard C++. However, My headers will be referred from Visual Studio 2015 as well as Visual Studio 2017 projects.

I would like to have something, such that for Visual Studio 2017 ( with C++ 17 lang feature set) , std::optional is used and with Visual Studio 2015, boost::optional gets used.

I am thinking of something like this:

#include <yvals.h>
#if _HAS_CXX17
 #include <optional>
 template <typename T> using Optional = std::optional<T>;
#else
 #include "boost/optional/optional.hpp"
 template <typename T> using Optional = boost::optional<T>;
#endif

Is it okay to use '_HAS_CXX17' macro this way? Is there a better way of doing this?

Answer 1

The short answer is: No It's not safe to rely on internal preprocessor defines in the implementation of the Visual C++ Runtime, and technically all compiler symbols that begin with a single _ are reserved for use by the implementation.

For example, _NOEXCEPT has been used internally in Visual Studio 2015 and 2017, but as of VS 2017 (15.8 update), this macro no longer exists; the headers just use noexcept directly.

The recommendation to use __has_include is good, but isn't supported prior to VS 2017 (15.3 update).

The other challenge is that __cplusplus doesn't indicate you are using /std:c++17 unless you are using VS 2017 (15.7 update) with the new /Zc:__cplusplus switch which is off by default.

Probably the safest way to do this across a range of VS versions would be:

#if (__cplusplus >= 201703L) || (defined(_MSVC_LANG) && (_MSVC_LANG >= 201703L) && (_MSC_VER >= 1913))
#if __has_include(<optional>)
 #include <optional>
 template <typename T> using Optional = std::optional<T>;
#else
 #include "boost/optional/optional.hpp"
 template <typename T> using Optional = boost::optional<T>;
#endif
#else
 #include "boost/optional/optional.hpp"
 template <typename T> using Optional = boost::optional<T>;
#endif

See Visual C++ Language Conformance