Re: An overloaded operator& needs the address of its argument

=?iso-8859-1?q?Erik_Wikstr=F6m?= <>
9 May 2007 23:58:36 -0700
On 10 Maj, 07:02, "Jim Langston" <> wrote:

"Sylvester Hesp" <> wrote in message


"Angel Tsankov" <> wrote in message

Angel Tsankov
"Sylvester Hesp" <> wrote in message

"Angel Tsankov" <> wrote in message

How can an overloaded operator& take the address of its argument:

template<typename T>
Smth operator &(T& SomeObject)
   // The address of SomeObject is needed here

It's argument is always 'this', as you can't define the unary & as a
non-member. And since 'this' is a pointer, you already have it's addr=



Does the standard say that unary address-of operator must be a member?=


so, where?

You're absolutely right, I was mistaken.
You could take the address by using a reinterpret_cast to a primitive t=


on which the unary & does what you want. boost::addressof does it like

template<class T> T* addressof(T& t)
   return reinterpret_cast<T*>(&const_cast<char&>(reinterpret_cast<const
volatile char&>(t)));

I must be missing something. Why wouldn't

 template<class T> T* addressof(T& t)
   return &*t;


First you dereference t (which means that T must either be a pointer
of implement operator *) and then you take the address of what was
returned. So if T was a normal pointer then you would return a copy of
t right?

However since the return-type is T* this does not compile for normal
pointers, nor for builtin functions. The only thing I can see this
working for is something like this:

struct Foo {
    Foo& operator*() {return *this;}

I think you must have forgotten something in your previous post.

Erik Wikstr=F6m

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