如何复制boost::u存储对象中的数据?

boost::aligned_storage 数据类型对我很有用,以便在c++11之前的世界中提供一致的存储。我有一个包含此存储成员的类:

template <size_t StoreSize>
class RoutineStorage {
   enum { ROUTINE_STORAGE_SIZE = StoreSize};
   enum { BUFFER_ALIGNMENT_VALUE = 8 };

   template <typename TStorageType> TStorageType& getStorageAsType()
   {
       BOOST_STATIC_ASSERT_MSG(boost::has_trivial_assign<TStorageType>::value &&
         boost::has_trivial_copy<TStorageType>::value,
         "The storage type must be trvially copyable and assignable to support this classes "
          "copy|assign semantics.");
        ... // Checking and some other code.
        return new (_store.address()) TStorageType();
   }

private:
    typedef boost::aligned_storage<ROUTINE_STORAGE_SIZE, BUFFER_ALIGNMENT_VALUE> 
        StorageBuffer;

    StorageBuffer _store;
}

我想为这个类提供一个复制构造函数,但是当我看到 aligned_storage 它有一个列为private的复制构造函数和一个注释 // noncopyable

RoutineStorage(const RoutineStorage<StoreSize>& copy)
{
    std::memcpy(_store.address(), copy._store.address(), _store.size())
}

这会有问题吗?据我所知 aligned_buffer address 函数将给出一个连续的内存地址和 size