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

James Kanze <>
Sun, 15 Jun 2008 13:10:17 -0700 (PDT)
On Jun 15, 9:24 pm, Paavo Helde <> wrote:

James Kanze <> kirjutas:

On Jun 14, 9:36 pm, Paavo Helde <> wrote:

"" <> 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.

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


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:

       / | \
 Derived1 Derived2 \
      \ | /

His code implements:

    Base Base
      | |
    Derived1 Derived2 Base
          \ | /

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

It seems Comeau online has the same opinion:

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

(In addition to mention an error regarding the incompatible throw

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)
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