friend 'operator new' in inheritance as a template
 

#include <iostream>
#include <map>
#include <utility>

//
// Base
// / | \
// Derived1 Derived2 \
// \ | /
// Derived3
//

template< typename _T >
struct Base {
friend void * operator new ( size_t _size ){
std::cout << typeid( _T ).name() << std::endl;
return malloc( _size );
}
};

struct Derived1 : Base< Derived1 > {
};

struct Derived2 : Base< Derived2 >{
};

struct Derived3 : Derived1, Derived2, Base< Derived3 >{
};

int main(){
Derived1 * d1 = new Derived1; // prints 8Derived3
Derived2 * d2 = new Derived2; // prints 8Derived3
}

Re: friend 'operator new' in inheritance as a template
 

I am sorry, with copy/paste I clear message

May anyony explain me how `operator new' templates argument is deduced
as Derived3?

Re: friend 'operator new' in inheritance as a template
 

Paavo Helde wrote:
> "wo3kie@gmail.com" <wo3kie@gmail.com> kirjutas:
>> template< typename _T >
>
> Undefined behavior, you are not allowed to use such identifiers in your
> code (beginning underscore followed by a capital letter). However, this
> is probably not the reason of your concern.
I thought identifiers could start with underscore (or even be
underscore), why is it undefined?

--
Daniel Pitts' Tech Blog: <http://virtualinfinity.net/wordpress/>

Re: friend 'operator new' in inheritance as a template
 

On 2008-06-14 22:28, Daniel Pitts wrote:
> Paavo Helde wrote:
>> "wo3kie@gmail.com" <wo3kie@gmail.com> kirjutas:
>>> template< typename _T >
>>
>> Undefined behavior, you are not allowed to use such identifiers in your
>> code (beginning underscore followed by a capital letter). However, this
>> is probably not the reason of your concern.
> I thought identifiers could start with underscore (or even be
> underscore), why is it undefined?

Identifiers starting with an underscore followed by a capital letter are
reserved for the implementation, also identifiers in the global
namespace beginning with an underscore are also reserved (so you can
only use them inside functions, classes, and namespaces.

--
Erik Wikström

Re: friend 'operator new' in inheritance as a template
 

On Jun 14, 9:36 pm, Paavo Helde <nob...@ebi.ee> wrote:
> "wo3...@gmail.com" <wo3...@gmail.com> kirjutas:

> > #include <iostream>
> > #include <map>
> > #include <utility>

> > //
> > // Base
> > // / | \
> > // Derived1 Derived2 \
> > // \ | /
> > // Derived3
> > //

> > template< typename _T >
> > struct Base {
> > friend void * operator new ( size_t _size ){
> > std::cout << typeid( _T ).name() << std::endl;

> As far as I understand, operator new cannot be a template, and
> a friend declaration does not make it into a template
> automatically anyway, so the _T symbol should not be visible
> inside the function. If the compiler still compiles this, I
> think this is a compiler bug.

The whole thing is fishy. This basically says that 1) the
global operator new is a friend of Base, and 2) provides an
implementation of the global operator new. Which is, I guess,
legal. Once... because this is a template, it's going to
provide a new implementation of the global operator new each
time the template is instantiated. Which is a violation of the
one definition rule, and so undefined behavior. (Since the
function is defined *in* the template, I'm pretty sure that the
use of _T is legal.)

There's also the problem that because the function is defined in
a class definition, it is inline; the global operator new
function is not allowed to be inline (for the obvious reason
that it's likely to have been used in some library code
somewhere, and that code won't see your inline definition).

> > return malloc( _size );

Which creates a couple of additional problems: if malloc fails,
his operator new is going to return a null pointer, rather than
raising a bad_alloc exception. And he hasn't provided a
corresponding operator delete (and of course, he has no idea as
to how the standard operator delete will try to free the
memory).

> > }
> > };

> > struct Derived1 : Base< Derived1 > {
> > };

> > struct Derived2 : Base< Derived2 >{
> > };

> > struct Derived3 : Derived1, Derived2, Base< Derived3 >{
> > };

Note that the code doesn't correspond to the diagram in comments
at the top.

> > int main(){
> > Derived1 * d1 = new Derived1; // prints 8Derived3
> > Derived2 * d2 = new Derived2; // prints 8Derived3
> > }

> You have provided illegal source to the compiler; the results
> can be whatever. I am not sure if the compiler is obliged to
> produce a diagnostic; I suspect it is. Maybe you should file a
> bug report to the compiler provider.

It's undefined behavior, but I can guess what is happening.
He's instantiated the template three times, which provides three
different implementations of the global operator new. The
compiler just happened to use the last one it saw.

--
James Kanze (GABI Software) email:james.kanze@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34

Re: friend 'operator new' in inheritance as a template
 

On Jun 15, 9:24 pm, Paavo Helde <nob...@ebi.ee> wrote:
> James Kanze <james.ka...@gmail.com> kirjutas:
> > On Jun 14, 9:36 pm, Paavo Helde <nob...@ebi.ee> wrote:
> >> "wo3...@gmail.com" <wo3...@gmail.com> kirjutas:

> >> > #include <iostream>
> >> > #include <map>
> >> > #include <utility>

> >> > //
> >> > // Base
> >> > // / | \
> >> > // Derived1 Derived2 \
> >> > // \ | /
> >> > // Derived3
> >> > //

> >> > template< typename _T >
> >> > struct Base {
> >> > friend void * operator new ( size_t _size ){
> >> > std::cout << typeid( _T ).name() << std::endl;

> >> As far as I understand, operator new cannot be a template, and
> >> a friend declaration does not make it into a template
> >> automatically anyway, so the _T symbol should not be visible
> >> inside the function. If the compiler still compiles this, I
> >> think this is a compiler bug.

> > The whole thing is fishy. This basically says that 1) the
> > global operator new is a friend of Base, and 2) provides an
> > implementation of the global operator new. Which is, I guess,
> > legal. Once... because this is a template, it's going to
> > provide a new implementation of the global operator new each
> > time the template is instantiated. Which is a violation of the
> > one definition rule, and so undefined behavior. (Since the
> > function is defined *in* the template, I'm pretty sure that the
> > use of _T is legal.)

> Yes, I somehow assumed that the definition of the friend function should
> not depend on whether it is "inline" in the friend declaration or not.
> And if it is not, then _T can't be visible in the function body. And _T
> is not participating in the function signature, so it would be impossible
> to forward this to the non-inline non-template function definition.
> right?

> But anyway, as Comeau online does not complain about use of _T (renamed
> to T to be sure), it seems my assumption was wrong. So the meaning of the
> function body can differ depending on if it is defined inline in the
> friend declaration, or separately. You learn something new every day...

What's legal in the inline definition of a friend doesn't depend
on the signature of the friend. And things like:

template< typename Derived >
class ArithmeticOperators
{
friend Derived
operator+( Derived const& lhs, Derived const& rhs )
{
Derived result( lhs ) ;
result += rhs ;
return result ;
}
// ...
} ;

Are more or less standard practice, and have been since Barton
and Nackman invented the idiom (or even before that, if they
learned it from somewhere else). (In this case, of course,
there's no problem with the one definition rule, because each
instantiation of ArithmeticOperators will define a different
operator+.)

> [...]

> >> > struct Derived1 : Base< Derived1 > {
> >> > };

> >> > struct Derived2 : Base< Derived2 >{
> >> > };

> >> > struct Derived3 : Derived1, Derived2, Base< Derived3 >{
> >> > };

> > Note that the code doesn't correspond to the diagram in comments
> > at the top.

> How is that? I can't see anything wrong here.

His diagram is:

Base
/ | \
Derived1 Derived2 \
\ | /
Derived3

His code implements:

Base Base
| |
Derived1 Derived2 Base
\ | /
Derived3

Not at all the same thing.

> >> > int main(){
> >> > Derived1 * d1 = new Derived1; // prints 8Derived3
> >> > Derived2 * d2 = new Derived2; // prints 8Derived3
> >> > }

> >> You have provided illegal source to the compiler; the results
> >> can be whatever. I am not sure if the compiler is obliged to
> >> produce a diagnostic; I suspect it is. Maybe you should file a
> >> bug report to the compiler provider.

> > It's undefined behavior, but I can guess what is happening.
> > He's instantiated the template three times, which provides three
> > different implementations of the global operator new. The
> > compiler just happened to use the last one it saw.

> Yes, you are right, this appears to be what gcc is doing. It
> seems VC++ 9.0 detects the multiple definitions error:

> error C2084: function 'void *Base<T>::operator new(size_t)' already has a
> body

> It seems Comeau online has the same opinion:

> "ComeauTest.c", line 15: error: function "operator new(size_t)" has
> already been
> defined
> friend void * operator new ( size_t _size ){
> ^
> detected during instantiation of class "Base<T> [with
> T=Derived2]"

> (In addition to mention an error regarding the incompatible throw
> specification).

Well, formally, it's undefined behavior (I think), so g++ isn't
wrong. But frankly, in this case, I'd expect an error message
as well.

--
James Kanze (GABI Software) email:james.kanze@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34