Re: Virtual Placement Delete?

From:
blargg.ei3@gishpuppy.com (blargg)
Newsgroups:
comp.lang.c++
Date:
Thu, 07 May 2009 17:20:42 -0500
Message-ID:
<blargg.ei3-0705091720420001@192.168.1.4>
Andrew Tomazos wrote:

On May 7, 11:32 pm, Victor Bazarov <v.Abaza...@comAcast.net> wrote:

[struct B has virtual dtor, and D is derived from B]

        void f()
        {
                void* p = malloc(sizeof(D));
                new (p) D; // ...placement new
                B* b = (B*) p;


That's *not* a good idea. You should do

            D* pd = new (p) D; // ...placement new
            B* b = pd;

instead.

                b->~B(); // ...is virtual destructor in effect?
                free(p);
        }

When f() is executed will D::~D() be called?


With your current cast from 'p' to a B*, the behaviour is undefined I'm
afraid.


(1) Can you tell me what the difference is between:

         new (p) D;
         B* b = (B*) p;

and

         D* pd = new (p) D;
         B* b = pd;

I would have thought they were equivalent? Why aren't they?


They are equivalent if p is of type D* AND you're using the usual global
placement new. If not, then you're using a void* in the first line to
point to a D, and in the second telling the compiler it points to a B. In
effect, you have this:

    void* p = new D;
    B* b = static_cast<B*> (p); // actually a fraudulent_cast<>

See a recent post where I elaborate on this (sorry for the web link):
<http://groups.google.com/group/comp.lang.c++/tree/browse_frm/thread/c4f204e6fc9067bd/37422027107e56d3?rnum=31#doc_86fad7521c2240dd>

(2) If I use your recommended form, will the virtual destructor of
D::~D be called if I call b->~B() ?


Yes.

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