Re: "moveable data type" in comparison with "r-value reference"

From:
"=?iso-8859-1?q?Pedro_Lamar=E3o?=" <pedro.lamarao@gmail.com>
Newsgroups:
comp.std.c++
Date:
Thu, 22 Mar 2007 17:34:27 CST
Message-ID:
<1174597584.373015.245540@e1g2000hsg.googlegroups.com>
On 21 mar, 20:33, grizl...@yandex.ru ("Grizlyk") wrote:

The state of the source may be unchanged, or it may be radically
different. The only requirement is that the object remain in a self
consistent state (all internal invariants are still intact).


The requirement is theoretically correct, but practically useless one,
because even auto_ptr wrapper has other behaviour.

Note, it is very important for auto_ptr that its state after "move" became
incorrect and user must not have access to auto_ptr during incorrect state,
because else we can not implement desired behaviour of auto memory
substitution.


Observe the implementation of the "pseudo-move constructor" for
auto_ptr in current libstdc++ main line:

      auto_ptr(auto_ptr& __a) throw() : _M_ptr(__a.release()) { }

      element_type*
      release() throw()
      {
    element_type* __tmp = _M_ptr;
    _M_ptr = 0;
    return __tmp;
      }

After we move from __a it becomes just as if default constructed.
So it is not true that move semantics require the moved-from object to
have incorrect internal state.

Article "apperance of moveable concept" on my pagehttp://grizlyk1.narod.ru/cpp_new#14_1coming soon.

From a client code point of view, choosing move instead of copy means
that you don't care what happens to the state of the source.


Here "copying from non-const source" and "move" are mixed. There are
differences beetwen them. For "move" source state _never_ will be correct or
UB, _always_ will be incorrect.


This _never_ is artificial.
There is no practical requirement for it.
This _never_ implies the above implementation of auto_ptr is wrong.

It is not.

--
 Pedro Lamar?o

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