Address-of operator

[Stub]

What does it mean that "C++ silently write and call address-of operators"?

[Andrey Tarasevich]

Re: Address-of operator

Stub wrote:[color=blue]
> What does it mean that "C++ silently write and call address-of operators"?[/color]

It doesn't seem to mean anything. Why don't you provide the complete
context for this fragment?

--
Best regards,
Andrey Tarasevich

[Nilesh]

Re: Address-of operator

"Stub" <stub@asof.co m> wrote in message news:<yXmqb.290 24$Ec1.2667822@ bgtnsc05-news.ops.worldn et.att.net>...[color=blue]
> What does it mean that "C++ silently write and call address-of operators"?[/color]

It means that you do not have to explicitally write 'operator &', just
as you dont have to write 'operator ='. C++ will silently provide its
own implementation.

Regards,
Nilesh Dhakras

[Ron Natalie]

Re: Address-of operator


"Nilesh" <dhakras_n@hotm ail.com> wrote in message news:ae9adc7c.0 311060352.2b7d8 e21@posting.goo gle.com...[color=blue]
> "Stub" <stub@asof.co m> wrote in message news:<yXmqb.290 24$Ec1.2667822@ bgtnsc05-news.ops.worldn et.att.net>...[color=green]
> > What does it mean that "C++ silently write and call address-of operators"?[/color]
>
> It means that you do not have to explicitally write 'operator &', just
> as you dont have to write 'operator ='. C++ will silently provide its
> own implementation.
>[/color]
Actually that is not true. No implementation is provided by default.
& is just one of those operators that has a native meaning when not
overloaded. This is distinct from the copy assignment operator which
when not declared is implicitly generated.

This is a subtle distinction but examine the following program:

#include <iostream>
using namespace std;

struct B {
B* operator&() {
cout << "B::operator&() \n";
return this;
}
};

struct D : B {
};

int main() {
D d;
&d;
return 0;
}

In this case the &d expression invokes the inheritted B::operator&. If there was an implicitly
generated one (as C++ does for operator=), then the one in B would be hidden and not used.

[Andrey Tarasevich]

Re: Address-of operator


Nilesh wrote:[color=blue]
> "Stub" <stub@asof.co m> wrote in message news:<yXmqb.290 24$Ec1.2667822@ bgtnsc05-news.ops.worldn et.att.net>...[color=green]
>> What does it mean that "C++ silently write and call address-of operators"?[/color]
>
> It means that you do not have to explicitally write 'operator &', just
> as you dont have to write 'operator ='. C++ will silently provide its
> own implementation.[/color]

No, there's absolutely nothing in common between copy assignment
operator and operator '&' in this respect. C++ does not "provide its own
implementation" for member operator &, silently or not.

--
Best regards,
Andrey Tarasevich

[wogston]

Re: Address-of operator

> No, there's absolutely nothing in common between copy assignment[color=blue]
> operator and operator '&' in this respect. C++ does not "provide its own
> implementation" for member operator &, silently or not.[/color]

I thought it does, unless you define own = operator, copy the members using
their respective = operators. What is the correct wording of this? Seems
that is the only thing you disagree about as far as I can tell.

-w-

[Andrey Tarasevich]

Re: Address-of operator

wogston wrote:
[color=blue][color=green]
>> No, there's absolutely nothing in common between copy assignment
>> operator and operator '&' in this respect. C++ does not "provide its own
>> implementation" for member operator &, silently or not.[/color]
>
> I thought it does, unless you define own = operator, copy the members using
> their respective = operators. What is the correct wording of this? Seems
> that is the only thing you disagree about as far as I can tell.
> ...[/color]

I'm talking about 'operator &'. Once again, C++ does not "provide its
own implementation" for member 'operator &', silently or not.

As for the copy assignment operator, you are right. If you don't declare
one in your class, the compiler will implicitly declare one for you (and
define it, if necessary and possible).

--
Best regards,
Andrey Tarasevich

[Nilesh]

Re: Address-of operator

"Ron Natalie" <ron@sensor.com > wrote in message news:<3faa65c6$ 0$75676$9a6e19e a@news.newshost ing.com>...[color=blue]
> "Nilesh" <dhakras_n@hotm ail.com> wrote in message news:ae9adc7c.0 311060352.2b7d8 e21@posting.goo gle.com...[color=green]
> > "Stub" <stub@asof.co m> wrote in message news:<yXmqb.290 24$Ec1.2667822@ bgtnsc05-news.ops.worldn et.att.net>...[color=darkred]
> > > What does it mean that "C++ silently write and call address-of operators"?[/color]
> >
> > It means that you do not have to explicitally write 'operator &', just
> > as you dont have to write 'operator ='. C++ will silently provide its
> > own implementation.
> >[/color]
> Actually that is not true. No implementation is provided by default.
> & is just one of those operators that has a native meaning when not
> overloaded. This is distinct from the copy assignment operator which
> when not declared is implicitly generated.
>[/color]

Thanks Ron for this clarification.

Nilesh Dhakras.