Re: Interaction of delete and destructor

From:

Mark P <usenet@fall2005REMOVE.fastmailCAPS.fm>

Newsgroups:

comp.lang.c++

Date:

Thu, 22 Jun 2006 21:30:59 GMT

Message-ID:

<n_Dmg.48967$fb2.5633@newssvr27.news.prodigy.net>

junw2000@gmail.com wrote:

I use the code below to study delete and destructor.

#include <iostream>
using namespace std;

struct A {
  virtual ~A() { cout << "~A()" << endl; }; //LINE1

  void operator delete(void* p) {
    cout << "A::operator delete" << endl;
    free(p);
  }
};

struct B : A {
  void operator delete(void* p) {
    cout << "B::operator delete" << endl; //LINE2
    free(p);
  }
  ~B(){ cout << "~B()" << endl; }; //LINE3
};

int main() {
  A* ap ;
  B *bp = new B;
  ap = bp;
  delete ap; //LINE4
}

The output of the code is:
~B()
~A()
B::operator delete

I can not understand the output. When LINE4 is executed, LINE1 is
executed, right?
Since ~A() is virtual, LINE3 is called. How can LINE2 be executed?
"delete ap" calls A's destructor. Then B's destructor is called. Can
B's destructor call B's operator delete?

In general, delete invokes destructor, right? Can destructor invoke
delete?

Thanks a lot. I can not find the answer from my C++ books.

Jack

Just a reminder that the "delete operator," the built-in operator that
you use when you write "delete [some_pointer];" is not the same as
"operator delete," the function that you may define within your class.
Conceptually you can think of the delete operator as doing two things:
invoking the destructor(s) of the object and then invoking operator
delete on the memory where the deconstructed object once was.

The distinction between the built-in "new operator" and a possibly
user-defined "operator new" is similar. When you write new [some_type],
thereby invoking the new operator, two things happen: operator new is
invoked which secures the memory for the object-to-be, and then the
constructor(s) is(are) invoked to actually construct the object.

Mark