此回答基于
@Johannes的答案/博客中展示的确切概念,因为这似乎是唯一“合法”的方法。我将该示例代码转换成了一个方便的实用程序。它易于与C++03兼容(通过实现
std::remove_reference
并替换
nullptr
)。
库
#define CONCATE_(X, Y) X##Y
#define CONCATE(X, Y) CONCATE_(X, Y)
#define ALLOW_ACCESS(CLASS, MEMBER, ...) \
template<typename Only, __VA_ARGS__ CLASS::*Member> \
struct CONCATE(MEMBER, __LINE__) { friend __VA_ARGS__ CLASS::*Access(Only*) { return Member; } }; \
template<typename> struct Only_##MEMBER; \
template<> struct Only_##MEMBER<CLASS> { friend __VA_ARGS__ CLASS::*Access(Only_##MEMBER<CLASS>*); }; \
template struct CONCATE(MEMBER, __LINE__)<Only_##MEMBER<CLASS>, &CLASS::MEMBER>
#define ACCESS(OBJECT, MEMBER) \
(OBJECT).*Access((Only_##MEMBER<std::remove_reference<decltype(OBJECT)>::type>*)nullptr)
API
ALLOW_ACCESS(<class>, <member>, <type>);
使用方法
ACCESS(<object>, <member>) = <value>
auto& ref = ACCESS(<object>, <member>)
struct X {
int get_member () const { return member; };
private:
int member = 0;
};
ALLOW_ACCESS(X, member, int);
int main() {
X x;
ACCESS(x, member) = 42;
std::cout << "proof: " << x.get_member() << std::endl;
}