有人能帮我讲一下这段构造函数是什么意思吗

我想要拷贝一份现有的NumericVector<Number>
下面一段是NumericVector的构造函数：

  /**
   * This _looks_ like a copy assignment operator, but note that,
   * unlike normal copy assignment operators, it is pure virtual. This
   * function should be overridden in derived classes so that they can
   * be copied correctly via references to the base class. This design
   * usually isn't a good idea in general, but in this context it
   * works because we usually don't have a mix of different kinds of
   * NumericVectors active in the library at a single time.
   *
   * \returns A reference to *this as the base type.
   */
  virtual NumericVector<T> & operator= (const NumericVector<T> & v) = 0;

  /**
   * The 5 special functions can be defaulted for this class, as it
   * does not manage any memory itself.
   */
  NumericVector (NumericVector &&) = default;
  NumericVector (const NumericVector &) = default;
  NumericVector & operator= (NumericVector &&) = default;

我用的另一个函数：

/**
 * Get a raw NumericVector with the given name.
 */
NumericVector<Number> &
SystemBase::getVector(const std::string & name)
{
  return system().get_vector(name);
}

然后我打的代码是：

NumericVector<Number> old_d = _nl.getVector("d");

我是想把_nl.getVector("d")拷贝出来放到old_d里面，但是报错了：

/home/user01/projects/moose/framework/src/executioners/Transient.C:471:54: error: cannot allocate an object of abstract type 'libMesh::NumericVector<double>'
  471 |       NumericVector<Number> old_d = _nl.getVector("d");

我看了下是因为我构造函数用的不对。
我有几个不懂：
1、=default 是不是代表 相当于我们常用的构造函数的语法，而不是什么都不做的一个空函数？
2、NumericVector && 是什么意思？
3、为什么构造函数可以用NumericVector &&和NumericVector &作为参数，不应该是NumericVector && a和NumericVector & a 之类的吗？
4、应该怎么拷贝？我试过

NumericVector<Number> old_d(_nl.getVector("d"));

但是也报错了