用 实现移动语义的正确方法是operator+
什么?类似于它的工作原理std::string
?
我尝试了以下方法,但是我希望有一些更优雅且可能更正确的方法来做到这一点:
class path
{
std::vector<std::string> path_;
public:
path& path::operator+=(const path& other)
{
path_.insert(std::begin(path_), std::begin(other.path_), std::end(other.path_));
return *this;
}
path& path::operator+=(path&& other)
{
path_.insert(std::begin(path_), std::make_move_iterator(std::begin(other.path_)), std::make_move_iterator(std::end(other.path_)));
return *this;
}
};
template<typename L, typename R>
typename std::enable_if<std::is_convertible<path, L>::value, path>::type operator+(const L& lhs, const R& rhs)
{
auto tmp = std::forward<L>(lhs);
tmp += std::forward<R>(rhs);
return tmp;
}