Seamlessly migrating experimental/optional to optional

From this Stack Overflow answer, I learned that C++17 will have __has_include, which can

[make] migrating from experimental to std almost seamless

This still leaves the question of how to do it. Doing something like namespace std { using namespace std::experimental; } is undefined behaviour, so I came up with this method:

optional.h demonstration on coliru

#pragma once  #if __has_include(<optional>) #   include <optional> #   define HAS_STD_OPTIONAL #elif __has_include(<experimental/optional>) #   include <experimental/optional> #   define HAS_STD_EXPERIMENTAL_OPTIONAL #else #   error Must have an optional type, either from <optional> or if not supported from <experimental/optional>. #endif  #if defined HAS_STD_OPTIONAL  namespace opt {     template<class T>     using optional = std::optional<T>;     using bad_optional_access = std::bad_optional_access;     using nullopt_t = std::nullopt_t;     using in_place_t = std::in_place_t;      constexpr auto nullopt = std::nullopt;     constexpr auto in_place = std::in_place;      template<class T>     constexpr auto make_optional(T && value)     {         return std::make_optional(std::forward<T>(value));     } }  #elif defined HAS_STD_EXPERIMENTAL_OPTIONAL  namespace opt {     template<class T>     using optional = std::experimental::optional<T>;     using bad_optional_access = std::experimental::bad_optional_access;     using nullopt_t = std::experimental::nullopt_t;     using in_place_t = std::experimental::in_place_t;      constexpr auto nullopt = std::experimental::nullopt;     constexpr auto in_place = std::experimental::in_place;      template<class T>     constexpr auto make_optional(T && value)     {         return std::experimental::make_optional(std::forward<T>(value));     } }  #endif 


Category: c# Time: 2016-07-29 Views: 0
Tags: c++17

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