Re: Deleting from destructor

From:
Rolf Magnus <ramagnus@t-online.de>
Newsgroups:
comp.lang.c++
Date:
Thu, 09 Oct 2008 21:59:14 +0200
Message-ID:
<gclnqi$viq$00$1@news.t-online.com>
Please don't top-post.

mc wrote:

"Rolf Magnus" <ramagnus@t-online.de> wrote in message
news:gclfpm$td3$01$1@news.t-online.com...

mc wrote:

Thanks Victor. I understand what you said and knew. Let me put more
context here but adding a more complete example:

void SKEL::bind(const MCU& mcu, ...)
{
    // const FOO& MCU::foo()
    // {
    // return (*new Foo());
    // }
    FOO foo = mcu.foo(); // Because of the const FOO&
returned, foo becomes equal to what was returned by MCU::foo()


Yes. It becomes a copy of it.

    // do stuff
    // when exising here, the destructor for FOO is called and the
    memory is release as per previous post


The memory for the object foo in SKEL::bind is, but there is the other
FOO object that was dynamically allocated. It's still hanging around, and
you lost all pointers to it, so you can't ever deallocate it. That's a
memory leak. With the destructor you described, you also happen to use
delete with
a pointer to memory you didn't get from new, which results in undefined
behavior.


I haven't lost the location where the object was; remember that in the
constructor a private member is initialized to the value of where the
object resides in memory; and the destructor uses a delete for that.


Yes, you're right. I missed that. So the copy's destructor frees the
original object's memory. However, the original's destructor isn't properly
called as it's supposed to, which probably won't lead to problems in the
example you showed, but will in less trivial programs. This is very bad
style.

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