copy constructor of std::auto_ptr<>

[dragoncoder]

Hi all,

I am trying to understanding std::auto_ptr<T class implementation from
"The C++ standard library" by Nicolai Josuttis. He gives a sample
implementation of auto_ptr class template in section 4.2. The copy
constructor is defined as:

auto_ptr (auto_ptr& rhs) throw()
: ap (rhs. release()) {
}

I was wondering, is there not a leak here ? I mean before taking
ownership of the memory pointed to by rhs, shouldn't the memory pointed
to by ap be relased (freed) ? I am expecting the copy constructor to be
this way:

auto_ptr(auto_p tr& rhs) throw() {
delete ap; // Isn't this required ?
ap = rhs.release();
}

If the delete ap is removed from the definition, I think the memory
which is already being pointed to by ap (before the copy constructor is
called) will leak. Please comment.

Thanks in advance.

/P

[Pete Becker]

Noah Roberts wrote:

Alf P. Steinbach wrote:

>* Salt_Peter:

>>copy ctors aren't called, they are invoked.

The C++ standard mostly uses the term "called". In a few places it uses
the term "invoked". There's no inherent difference in meaning, but no
matter whether "call" or "invoked" is used, there are several different
possible meanings, and the one that applies depends on the context.

Uh oh..not this one again :P

Actually the future standard is going to have a very specific
definition of invoke. See section 3 in TR1.

That's INVOKE, not invoke. C++ is case sensitive. INVOKE and INVOKE_R
are used to describe the effects of a function call operator applied to
any of the four callable types defined in TR1. They don't affect the
meaning of the English word 'invoke'.

--

-- Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com)
Author of "The Standard C++ Library Extensions: a Tutorial and
Reference." (www.petebecker.com/tr1book)