> Doesn't this mean that Mutex introduces one more pointer? No. That syntax is roughly equivalent to the following C++: auto const lock = std::make_shared

> Doesn't this mean that Mutex introduces one more pointer? No. That syntax is roughly equivalent to the following C++: auto const lock = std::make_shared<std::pair<std::mutex, uint32_t>>( std::piecewise_construct, std::make_tuple(), std::make_tuple(0));