Re: noob question - destructors - freeing memory...

From:
"MikeWhy" <boat042-nospam@yahoo.com>
Newsgroups:
comp.lang.c++
Date:
Fri, 30 Dec 2011 20:06:57 -0600
Message-ID:
<jdlqo9$i9j$1@dont-email.me>
someone wrote:

Hi

See this (stolen from the internet):

----------------------------------------
#include <iostream>
using namespace std;

class Base
{
public:
  Base(){ cout<<"Constructing Base"<<endl;}

  // this is a destructor:
  virtual ~Base(){ cout<<"Destroying Base"<<endl;}
};

class Derive: public Base
{
public:
  Derive(){ cout<<"Constructing Derive"<<endl;}
  ~Derive(){ cout<<"Destroying Derive"<<endl;}
};

int main()
{
  Base *basePtr = new Derive();
  // delete basePtr; // <<<======== !! ! ! ! ! NB !!
  return 0;
}

----------------------------------------

See middle line of main():

  delete basePtr; // <<<======== !! ! ! ! ! NB !!

If I enable (uncomment) middle line, I get:
Constructing Base
Constructing Derive
Destroying Derive
Destroying Base

If I disable (comment out) middle line, I get:
Constructing Base
Constructing Derive

========

This, I didn't quite expect, because I thought that at the end of the
program, all destructors where guaranteed to be called...


If pointers had destructors, what action should they take?

In my case, I have a longer program and I allocate memory with one of
my constructors, hence it is crucial to free the memory, but since the
virtual destructor is never called, I never free the memory...

In addition: In my full program (not shown here), I instantiate my new
object like:

int main(...)
{
  ...
  Derive d1();
  return 0; // or exit(1); if something went wrong somewhere
}

Is it required that I do, like in the example above:

int main(...)
{
  ...
  Base *myPtr = new Derive();


    foo(); // insert this line. See below.

  delete myPtr;
  return 0; // or exit(1); if something went wrong somewhere
}

???????


Yes.

But what if my code does an exit(1) somewhere, which it could do at
least a dozens of times, in different parts of the program?


exit() is a function call. In your final example above, if foo() called
exit(1), will your destructor be called? Given what you saw earlier, how
does exit() trigger process termination?

If instead of calling exit(), foo() throws an uncaught exception. How does
the program behave?

Calling exit() is a very bad idea. Ditto for terminating threads, only more
so. A simple return will suffice. Find some other mechanism to trigger
process (and thread) termination.

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