How do you think?
and why?
9  2245 
BekTek wrote: How do you think?
and why?
Boost has several different smart pointer types, each tailored for a
specific need, which allows for more flexibility. I would go with boost.
I have even heard that boost smart pointers are subject to enter the C++
standard library on the next revision.
- Matthias
In article <co************ *@news.t-online.com>,
Matthias Kappler <no****@digital raid.com> wrote: BekTek wrote:
How do you think?
and why?
Boost has several different smart pointer types, each tailored for a
specific need, which allows for more flexibility. I would go with boost.
I have even heard that boost smart pointers are subject to enter the C++
standard library on the next revision.
- Matthias
Proposals based on boost::shared_p tr and boost::weak_ptr have been
accepted into the first library technical report (TR1). This does not
mean that they are standard, only that they are being seriously
considered for standardization . Other boost smart pointers (scoped_ptr,
shared_array, intrusive_ptr, etc.) have not been proposed.
As to the OP's original question, it is not answerable without some
context on what the smart pointer is supposed to do. I would not advise
using shared_ptr when auto_ptr is a better fit, nor vice-versa.
-Howard
"BekTek" <be****@gmail.c om> wrote in message news:5Ehqd.3000 $xk5.755@trnddc 09... How do you think?
and why?
As others have mentioned, boost::shared_p tr and std::auto_ptr are useful in
different types of situations. If you decide that auto_ptr is more appropriate,
however, you might also consider this implementation of move_ptr (based in large
part on Howard's work): http://home.comcast.net/~jturkanis/move_ptr/
(It is used as an implementation detail in the recently reviewed Boost Smart
Containers library, by Thorsten Ottosen, and may therefore soon become an
undocumented part of Boost.)
The library documentation gives a detailed analysis of the differences between
move_ptr and auto_ptr, as well as some useful references on auto_ptr.
Jonathan
"BekTek" <be****@gmail.c om> wrote in message
news:5Ehqd.3000 $xk5.755@trnddc 09... How do you think?
and why?
They both have their uses, so I wouldn't call one "better". However,
std::auto_ptr does have a characteristic which is hard to like: for
auto_ptr's x and y
y = x;
changes both x and y. This can certainly cause confusion in use. A similar
propery of auto_ptr's copy constructor makes in unusable in standard
containers such as std::vector. Also note that a class which contains an
auto_ptr as a member will have the same dubioius properties.
boost::shared_p tr has none of these issues.
--
Cy http://home.rochester.rr.com/cyhome/
"Cy Edmunds" <ce******@spaml ess.rochester.r r.com> wrote in message
news:qj******** **********@twis ter.nyroc.rr.co m... "BekTek" <be****@gmail.c om> wrote in message
news:5Ehqd.3000 $xk5.755@trnddc 09... How do you think?
and why?
They both have their uses, so I wouldn't call one "better". However,
std::auto_ptr does have a characteristic which is hard to like: for
auto_ptr's x and y
y = x;
changes both x and y. This can certainly cause confusion in use.
This one of the main attractions of move_ptr: the above won't compile. If you
really want to transfer ownership from x to y, you have to write
y = move(x);
Jonathan
In article <30************ *@uni-berlin.de>,
"Jonathan Turkanis" <te******@kanga roologic.com> wrote: "Cy Edmunds" <ce******@spaml ess.rochester.r r.com> wrote in message
news:qj******** **********@twis ter.nyroc.rr.co m... "BekTek" <be****@gmail.c om> wrote in message
news:5Ehqd.3000 $xk5.755@trnddc 09... How do you think?
and why?
They both have their uses, so I wouldn't call one "better". However,
std::auto_ptr does have a characteristic which is hard to like: for
auto_ptr's x and y
y = x;
changes both x and y. This can certainly cause confusion in use.
This one of the main attractions of move_ptr: the above won't compile. If you
really want to transfer ownership from x to y, you have to write
y = move(x);
And kudos to Jonathan for implementing and making this available in
C++03 (as much as possible). I'm very much looking forward to the day
when this animal can be put into std::containers , and used with
std::algorithms . :-)
-Howard
Jonathan Turkanis wrote: "Cy Edmunds" <ce******@spaml ess.rochester.r r.com> wrote: for auto_ptr's x and y
y = x;
changes both x and y. This can certainly cause confusion in use.
This one of the main attractions of move_ptr: the above won't
compile. If you really want to transfer ownership from x to y, you have to write
y = move(x);
Apparently I have become seriously stupid but I don't see it:
| std::auto_ptr<i nt> x(new int(1));
| std::auto_ptr<i nt> y;
| y = x;
You seem to assert that the assignment in the above statement does not
compile. I can see that it may have unexpected behavior but not that it
does not compile. Can you please shed some light on this? Maybe you
were
refering to the situation where a function is returning an 'auto_ptr':
| std::auto_ptr<i nt> foo() { return std::auto_ptr<i nt>(new int(1)); }
| int main() {
| std::auto_ptr<i nt> x = foo(); // OK: there is a special ctor (*)
| x = foo(); // error: need non-const reference
| }
(*) At least, this is supposed to work. I seem to remember a discussion
which somehow deduced that this actually does not work. Is this
what you are refering to?
--
<mailto:di***** ******@yahoo.co m> <http://www.dietmar-kuehl.de/>
<http://www.contendix.c om> - Software Development & Consulting
In article <11************ *********@z14g2 000cwz.googlegr oups.com>,
"Dietmar Kuehl" <di***********@ yahoo.com> wrote: Jonathan Turkanis wrote: "Cy Edmunds" <ce******@spaml ess.rochester.r r.com> wrote: for auto_ptr's x and y
y = x;
changes both x and y. This can certainly cause confusion in use.
This one of the main attractions of move_ptr: the above won't
compile. If you really want to transfer ownership from x to y, you have to write
y = move(x);
Apparently I have become seriously stupid but I don't see it:
| std::auto_ptr<i nt> x(new int(1));
| std::auto_ptr<i nt> y;
| y = x;
You seem to assert that the assignment in the above statement does not
compile. I can see that it may have unexpected behavior but not that it
does not compile. Can you please shed some light on this? Maybe you
were
refering to the situation where a function is returning an 'auto_ptr':
| std::auto_ptr<i nt> foo() { return std::auto_ptr<i nt>(new int(1)); }
| int main() {
| std::auto_ptr<i nt> x = foo(); // OK: there is a special ctor (*)
| x = foo(); // error: need non-const reference
| }
Hi Dietmar,
What Jonathan meant was that if the type of the smart pointer is
move_ptr, instead of auto_ptr, then the assignment won't compile:
y = x; // compile time error
This is a safety feature of move_ptr, especially in generic code where
moving from lvalues with copy syntax can render the generic algorithm
incorrect.
-Howard
"Howard Hinnant" <hi*****@metrow erks.com> wrote: "Dietmar Kuehl" <di***********@ yahoo.com> wrote:
Hi Dietmar,
What Jonathan meant was that if the type of the smart pointer is
move_ptr, instead of auto_ptr, then the assignment won't compile:
y = x; // compile time error
This is a safety feature of move_ptr, especially in generic code where
moving from lvalues with copy syntax can render the generic algorithm
incorrect.
Yes, that's what I was trying to say. Sorry I wan't clear.
-Howard
Jonathan This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
by: SerGioGio |
last post by:
Hello !
When a ref. counted smart pointer (template) class is available (like
boost::shared_ptr), is there any (semantic) reason to still use
std::auto_ptr rather than this smart pointer ? Or should one use shared_ptr
for every cases ?
Thanks in advance for your remarks !
SerGioGio
|
by: ma740988 |
last post by:
Part of my confusion here is certainly my ignorance of templates and
std::auto_ptr. Two topics, I've perused but need to really _delve_
into. In any event, consider the 'test' source.
# include <iostream>
# include <memory>
class CallbackBase
{
public:
virtual void operator()() const { };
|
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Hi
Please try this code. I think that it's perfectly legal. However when
compiled under MSVC71 stack overflow happens in first line of main, thus
second line is never executed.
B.
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