Does std::optional copy an object containing a std::unique_ptr

Question

I'm trying to return a std::optional to an object containing a std::unique_ptr member. However, when I try to change the optional return type to use a reference, I get errors like this:

error: cannot bind non-const lvalue reference of type std::optional<ObjectWithConnection>& to an rvalue of type std::optional<ObjectWithConnection>

Is this the best way to return _aConnection without encountering any copying/latency across the stack frames?

I use references very often, but I've never used std::optional with std::unique_ptr and I'm concerned what it's doing under the hood.

struct ObjectWithConnection
{
    std::unique_ptr<Connection> _aConnection;
};

struct ConnectionsContainer
{
    // If I change return type to reference, get compiler error
    std::optional<ObjectWithConnection> getConnection(const std::string& key)
    {
        auto iter = _connections.find(key);
        if(iter != _connections.end())
        {
            return std::optional<ObjectWithConnection>(std::move(iter->second));
        }

        return std::nullopt;
    }

    std::unordered_map<std::string, ObjectWithConnection> _connections;
}

struct AnotherClass
{
    // If I change return type to reference, get compiler error
    std::optional<ObjectWithConnection> getConnection(const std::string& key)
    {
        return _container.getConnection(key);
    }

    ConnectionsContainer _container;
};

Answer 1

There are no std::optional references. A program is ill-formed if it instantiates an optional with a reference type. Alternatively, an optional of a std::reference_wrapper of type T may be used to hold a reference.