Re: std::atexit

From:
James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Wed, 2 Jan 2008 04:02:33 -0800 (PST)
Message-ID:
<dd4933ac-c2c2-4ad7-bdac-f012271f0e43@c4g2000hsg.googlegroups.com>
On Jan 2, 6:19 am, "Christopher Pisz" <some...@somewhere.net> wrote:

I am attempting to write a "Phoenix Singleton" using the book
"Modern C++ Design" by Alexandrescu I do not understand his
use of...

#ifndef ATEXIT_FIXED
std::atexit(Kill);
#endif

...in the source below. I understand the problem, but not how
a preprocessor directive will fix it. He says the standard is
unclear about the situation where one call to register with
std::atexit is the result of is made as an effect of another
std::atexit registration.


C99 is not unclear about this (although I seem to remember it
being undefined behavior in C90, and thus in C++98).

Can anyone be more specific on how to fix the problem?


Which problem? The code you posted has several different cases
of undefined behavior. All the #ifndef does is cause atexit not
to be called if ATEXIT_FIXED is defined.

I've not studied it in detail, but even after a quick glance, it
is apparent that if Kill() is ever called (and it will be called
if atexit is called), the destructor is called twice for the
same object. This is undefined behavior, and in a non-trivial
class, will almost certainly get you into trouble. So you
almost certainly have to defined ATEXIT_FIXED for the code to
work.

The obvious way to achieve the supposed goal (that an instance
will always be available) is to create the instance with new,
and never destruct it.

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