Re: What's the standard say about this code?

From:
Maxim Yegorushkin <maxim.yegorushkin@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Wed, 1 Oct 2008 08:11:46 -0700 (PDT)
Message-ID:
<fdf505e6-81d0-40d4-89cd-c9ad531e5a1d@j22g2000hsf.googlegroups.com>
On Oct 1, 12:49 pm, Triple-DES <DenPlettf...@gmail.com> wrote:

On 1 Okt, 12:50, "Daniel T." <danie...@earthlink.net> wrote:

James Kanze <james.ka...@gmail.com> wrote:

"Daniel T." <danie...@earthlink.net> wrote:

#include <cassert>

class Foo {
public:
        virtual void fnA() = 0;
        virtual void fnB() = 0;
};

int main() {
        assert( &Foo::fnB );
        assert( &Foo::fnA );
}

What does the standard say about the above code?


There should be no problem with it.

In the compiler I'm using now, the first assert will not fire,
but the second one will. I expected that neither assert would
fire...


It's guaranteed by the standard. It works with the three
compilers I have access to (Sun CC, g++ and VC++), at least when
you compile in standard conformant mode. (Note that by default,
pointers to member functions do not work in VC++. You must use
the option /vmg. Not that I think that their non-conformity
otherwise would play a role here.)


Can I get chapter and verse from the standard on this? It sounds like I
need to submit a bug report to the compiler vender.


I believe it is disallowed by 4.11/1 [conv.mem]:

"A null pointer constant (4.10) can be converted to a pointer to
member type; the result is the null member pointer value of that type
and is distinguishable from any pointer to member not created from a
null pointer constant.(...)"

Therefore, an rvalue obtained by using address-of on a member shall
not yield a null pointer constant.


In the original question a member function pointer gets converted to
bool. Is 4.11/1 applicable here?

--
Max

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