Re: Constructing Derived in shell of Base <shudder>

From:

Kai-Uwe Bux <jkherciueh@gmx.net>

Newsgroups:

comp.lang.c++

Date:

Wed, 14 Jul 2010 10:32:23 +0200

Message-ID:

<i1jsmo$rg8$1@news.doubleSlash.org>

Victor Bazarov wrote:

On 7/13/2010 4:17 PM, Kai-Uwe Bux wrote:

For the code

struct Base { int j; virtual void f(); };
struct Derived : public Base { virtual void f(); };
void fooBar
{
         Base b;
         b.f(); // Base::f() invoked
         b.~Base();
         new (&b) Derived; // placement new => as same size,
         construct Derived in shell of Base
         b.f(); // Which f() is invoked, Base or

Derived?

}

1) Is it well-defined? if not, why not?

It's not because the requirements of [3.8/7] are not met:

   [it works if]
   ...
   ? the new object is of the same type as the original object (ignoring
   the top-level cv-qualifiers), and
   ...

Note: there are other requirements, which I did not quote, that also need
to be satisfied.

Same type? Do you know why that is? The example in [3.8/7]
demonstrates those actions (calling the d-tor with the subsequent
placement new) inside a member function. Also, considering [3.8/8], the
B object is not destructed before T is constructed on top of it.

Not exactly, but here is a data point. The following "should" print

  ~B
  ~D
  ~B

#include <iostream>
#include <ostream>

struct Base {

  virtual
  ~Base ( void ) {
    std::cout << "~B\n";
  }

};

struct Derived : public Base {

  virtual
  ~Derived ( void ) {
    std::cout << "~D\n";
  }

};

int main ( void ) {
  Base b;
  b.~Base();
  new (&b) Derived;
}

But with g++, I get

~B
~B

The compiler bypasses a call via the vtable, presumably as it knows the type
of the object. As confirmation:

int main ( void ) {
  Base* b = new Base;
  b->~Base();
  new (b) Derived;
  delete b;
}

prints the expected results.

Best

Kai-Uwe Bux