Question

I would like to know standard’s view on dereferencing pointer to base, but I’m not making any progress finding it. Take these two classes for example:

class Base
{
  public:
    virtual void do_something() = 0;
};

class Derived : public Base
{
  public:
    virtual void do_something();
};

void foo2(Base *b)
{
     Base &b1 = *b; // how this work by standard?
}

void foo()
{
  Derived *d = new Derived();
  foo2(d); // does this work by standard?
}

So, basically, if pointer of type B to an object of type D is dereferenced, will slicing happen in place, or temporary will emerge? I’m prone to believe that temporary is not an option, because that would mean that temporary is instance of abstract class.

Whatever the truth, I would appreciate any pointers to the ISO standard that says one or the other. (Or third, for that matter. 🙂 )

EDIT:

I threw the point with temporary not being an option as a possible line of reasoning why it behaves the way it does, which is quite logical, but I can’t find confirmation in standard, and I’m not a regular reader.

EDIT2:

Through discussion, it became obvious that my question was actually about dereferencing a pointer mechanism, and not about splicing or temporaries. I thank everyone for trying to dumb it down for me, and I finally got answer to the question the puzzled me the most: Why I can’t find anything in the standard about this… Obviously it was the wrong question, but I’ve got the right answer.

Thnx

Answer 1

Base &b = *static_cast<Base *>(d); // does this work by standard?

Yes.

But you can simply do this:

Base &b = *d;

//use b polymorphically!
b.do_something(); //calls Derived::do_something()

No need to use static_cast. After all, Derived is derived from Base.

---

Reply to your edit:

foo2(d); // does this work by standard?

Yes. Pointer of type Base* can be initialized with pointer of type Derived*.

—

Base &b = *b; // how this work by standard?

No. They’re same name. If you mean, Base &b1 = *b, then yes, that works. b1 refers to the object pointed to by b.