How can I reliably get an object's address when operator& is overloaded?

Question

Consider the following program:

struct ghost
{
    // ghosts like to pretend that they don\'t exist
    ghost* operator&() const volatile { return 0; }
}         


        

Answer 1

The trick behind boost::addressof and the implementation provided by @Luc Danton relies on the magic of the reinterpret_cast; the standard explicitly states at §5.2.10 ¶10 that

An lvalue expression of type T1 can be cast to the type “reference to T2” if an expression of type “pointer to T1” can be explicitly converted to the type “pointer to T2” using a reinterpret_cast. That is, a reference cast reinterpret_cast<T&>(x) has the same effect as the conversion *reinterpret_cast<T*>(&x) with the built-in & and * operators. The result is an lvalue that refers to the same object as the source lvalue, but with a different type.

Now, this allows us to convert an arbitrary object reference to a char & (with a cv qualification if the reference is cv-qualified), because any pointer can be converted to a (possibly cv-qualified) char *. Now that we have a char &, the operator overloading on the object is no longer relevant, and we can obtain the address with the builtin & operator.

The boost implementation adds a few steps to work with cv-qualified objects: the first reinterpret_cast is done to const volatile char &, otherwise a plain char & cast wouldn't work for const and/or volatile references (reinterpret_cast cannot remove const). Then the const and volatile is removed with const_cast, the address is taken with &, and a final reinterpet_cast to the "correct" type is done.

The const_cast is needed to remove the const/volatile that could have been added to non-const/volatile references, but it does not "harm" what was a const/volatile reference in first place, because the final reinterpret_cast will re-add the cv-qualification if it was there in first place (reinterpret_cast cannot remove the const but can add it).

As for the rest of the code in addressof.hpp, it seems that most of it is for workarounds. The static inline T * f( T * v, int ) seems to be needed only for the Borland compiler, but its presence introduces the need for addr_impl_ref, otherwise pointer types would be caught by this second overload.

Edit: the various overloads have a different function, see @Matthieu M. excellent answer.

Well, I'm no longer sure of this either; I should further investigate that code, but now I'm cooking dinner :) , I'll have a look at it later.

Answer 2

I've seen an implementation of addressof do this:

char* start = &reinterpret_cast<char&>(clyde);
ghost* pointer_to_clyde = reinterpret_cast<ghost*>(start);

Don't ask me how conforming this is!

Answer 3

Take a look at boost::addressof and its implementation.

Answer 4

Update: in C++11, one may use std::addressof instead of boost::addressof.

---

Let us first copy the code from Boost, minus the compiler work around bits:

template<class T>
struct addr_impl_ref
{
  T & v_;

  inline addr_impl_ref( T & v ): v_( v ) {}
  inline operator T& () const { return v_; }

private:
  addr_impl_ref & operator=(const addr_impl_ref &);
};

template<class T>
struct addressof_impl
{
  static inline T * f( T & v, long ) {
    return reinterpret_cast<T*>(
        &const_cast<char&>(reinterpret_cast<const volatile char &>(v)));
  }

  static inline T * f( T * v, int ) { return v; }
};

template<class T>
T * addressof( T & v ) {
  return addressof_impl<T>::f( addr_impl_ref<T>( v ), 0 );
}

What happens if we pass a reference to function ?

Note: addressof cannot be used with a pointer to function

In C++ if void func(); is declared, then func is a reference to a function taking no argument and returning no result. This reference to a function can be trivially converted into a pointer to function -- from @Konstantin: According to 13.3.3.2 both T & and T * are indistinguishable for functions. The 1st one is an Identity conversion and the 2nd one is Function-to-Pointer conversion both having "Exact Match" rank (13.3.3.1.1 table 9).

The reference to function pass through addr_impl_ref, there is an ambiguity in the overload resolution for the choice of f, which is solved thanks to the dummy argument 0, which is an int first and could be promoted to a long (Integral Conversion).

Thus we simply returns the pointer.

What happens if we pass a type with a conversion operator ?

If the conversion operator yields a T* then we have an ambiguity: for f(T&,long) an Integral Promotion is required for the second argument while for f(T*,int) the conversion operator is called on the first (thanks to @litb)

That's when addr_impl_ref kicks in. The C++ Standard mandates that a conversion sequence may contain at most one user-defined conversion. By wrapping the type in addr_impl_ref and forcing the use of a conversion sequence already, we "disable" any conversion operator that the type comes with.

Thus the f(T&,long) overload is selected (and the Integral Promotion performed).

What happens for any other type ?

Thus the f(T&,long) overload is selected, because there the type does not match the T* parameter.

Note: from the remarks in the file regarding Borland compatibility, arrays do not decay to pointers, but are passed by reference.

What happens in this overload ?

We want to avoid applying operator& to the type, as it may have been overloaded.

The Standard guarantees that reinterpret_cast may be used for this work (see @Matteo Italia's answer: 5.2.10/10).

Boost adds some niceties with const and volatile qualifiers to avoid compiler warnings (and properly use a const_cast to remove them).

• Cast T& to char const volatile&
• Strip the const and volatile
• Apply the & operator to take the address
• Cast back to a T*

The const/volatile juggling is a bit of black magic, but it does simplify the work (rather than providing 4 overloads). Note that since T is unqualified, if we pass a ghost const&, then T* is ghost const*, thus the qualifiers have not really been lost.

EDIT: the pointer overload is used for pointer to functions, I amended the above explanation somewhat. I still do not understand why it is necessary though.

The following ideone output sums this up, somewhat.

Answer 5

Use std::addressof.

You can think of it as doing the following behind the scenes:

1. Reinterpret the object as a reference-to-char
2. Take the address of that (won’t call the overload)
3. Cast the pointer back to a pointer of your type.

Existing implementations (including Boost.Addressof) do exactly that, just taking additional care of const and volatile qualification.