Semantics of OBJ_TYPE_REF

[Erick Ochoa via Gcc]

Hi,

I am having some trouble understanding the semantics of OBJ_TYPE_REF.
I understand that it is made of three operands:

1. OBJ_TYPE_REF_EXPR: An expression that evaluates the value to use.
2. OBJ_TYPE_REF_OBJECT: Is the object on whose behalf the lookup is
being performed
3. OBJ_TYPE_REF_TOKEN: An integer index to the virtual method table.

The language here is very general and makes it hard for me to understand. Is

1. OBJ_TYPE_REF_EXPR: The virtual function
2. OBJ_TYPE_REF_OBJECT: The "this" object

?

Why is the index needed if it looks like the virtual function is
already pointed to by the first operand? I am reading Hubicka's post
on devirtualization and Part II
(https://hubicka.blogspot.com/2014/01/devirtualization-in-c-part-2-low-level.html)
seems to agree with me on the example below. There is also no mention
of OBJ_TYPE_REF on part II, but there is on part III.

Hubicka's example:

int test() ()
{
  int (*__vtbl_ptr_type) () *tmp1;
  int (*__vtbl_ptr_type) () tmp2;
  struct A a;
    struct A * b;


  _ZN1AC2Ev (&a);
  b = &a;

  tmp1 = b->_vptr.A;
  tmp2 = *tmp1;

  // Notice no OBJ_TYPE_REF below.
  return tmp2 (b);
}

My example:

#include <cstdio>

class A
{
public:
  virtual void foo() { printf("hello\n"); };
  virtual void bar() { printf("world\n"); };
};

int
main(int argc, char**argv )
{
        class A* a = new A();
        a->foo();
        a->bar();
        return 0;
}

// Relevant gimple:

    a = D.2922;
    _1 = a->_vptr.A;
    _2 = *_1; // first element of the virtual table?

    OBJ_TYPE_REF(_2;(struct A)a->0) (a);
    // wouldn't this just be equivalent to
    // _2 (a)

    _3 = a->_vptr.A;
    _4 = _3 + 8;

    _5 = *_4; // second element of the virtual table
    OBJ_TYPE_REF(_5;(struct A)a->1) (a);
    // same
    // _5 (a)

On part III Hubicka says that OBJ_TYPE_REF is a wrapper and that it
represents the type and token. (My guess is that these are the
arguments 2 and 3). I fail to understand why these are needed if we
have the function already.

Thanks! Any help is appreciated.