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

From:

Paavo Helde <nobody@ebi.ee>

Newsgroups:

comp.lang.c++

Date:

Sat, 14 Jun 2008 14:36:33 -0500

Message-ID:

<Xns9ABDE5FEE58D1nobodyebiee@216.196.97.131>

"wo3kie@gmail.com" <wo3kie@gmail.com> kirjutas:

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

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

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.

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.

        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
}

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.

Regards
Paavo