Can Pointer to reference and Pointer to the actual variable be treated as same?

Question

I recently saw the below code ( simplified version given below) in my code base and got this doubt:

class B;
class A
{
  public:
     A():m_A("testA"){}
     B& getB()
     {
       return m_B;
     }
     B* getBPtr() //== > written to explain the problem clearly
     {
       return &m_B;
     }
private:
    B m_B;
};

class B
{
  public:
    B(const std::string& name):m_Name(name){}
    std::string getName() const
    {
      return m_Name;
    }
private:
    std::string m_Name;
};

class C
{
 public:
   void testFunc(B* ptr)
   {
   }
};


int main()
{
  A a;
  C c;
 c.testFunc(&a.getB()); ===> is it equivalent to c.testFunc(a.getBPtr()) ?
}

1. The pointer to reference and pointer to actual variable can be treated as same?
2. Does standard says anything on the address of reference to be interchangeable used for address of variable.

Answer 1

Yes, it's not possible to take the address of a reference. You always get the address of the referred object.

Answer 2

Yes it is equivalent. Taking the address of a reference returns the address of the original thing. Since there is no way to get any other address out of a reference, there is no seperate type for a pointer to a reference.

Answer 3

First sentence of standard 8.3.2/4:

There shall be no references to references, no arrays of references, and no pointers to references

(My emphasis)

This doesn't mean you can't take the address of a variable declared as a reference, it just means that there's no separate type for a pointer to a reference.