cast to non-const reference of a function's return object

[klaus triendl]

hi,

recently i discovered a memory leak in our code; after some investigation i
could reduce it to the following problem:
return objects of functions are handled as temporary objects, hence their
dtor is called immediately and not at the end of the function. to be able to
use return objects (to avoid copying) i often assign them to a const
reference.
now, casting a const return object from a function to a non-const reference
to this return object calls immediately the dtor of the return object
anyway, any further operation deals with a non-valid object then. if i do
this in two steps - first holding a const reference to the return object and
then const_casting it, everything works like i expected it.

does anybody know whether the compiler behaves correctly?
i use the following short example to illustrate my question:

const string getastring()
{
return string();
}

void funcWmemleak()
{
string& str = const_cast<string&>(getastring());
str = "test"; // this is already an invalid object because the dtor was
called before (and won't ever be called again of course)
}

void funcWOmemleak()
{
const string& str1 = getastring();
string& str = const_cast<string&>(str1);
str = "test";
}

int main()
{
funcWmemleak();
funcWOmemleak();
}

--
klaus triendl

[tom_usenet]

On Wed, 02 Jun 2004 08:09:01 GMT, "klaus triendl" <tr********@mbox.at>
wrote:

hi,

recently i discovered a memory leak in our code;
I think you mean "use of a dangling reference" rather than "memory
leak".

after some investigation icould reduce it to the following problem:
return objects of functions are handled as temporary objects, hence their
dtor is called immediately and not at the end of the function. to be able to
use return objects (to avoid copying) i often assign them to a const
reference.
now, casting a const return object from a function to a non-const reference
to this return object calls immediately the dtor of the return object
anyway, any further operation deals with a non-valid object then. if i do
this in two steps - first holding a const reference to the return object and
then const_casting it, everything works like i expected it.

does anybody know whether the compiler behaves correctly?

Yes, binding a temporary directly to a const reference extends the
lifetime of the temporary to match that of the reference. You can't
bind a temporary to a non-const reference, so there's no way of doing
the lifetime extension using a non-const ref.

Note, since you return a const string, using the returned string as a
non-const string results in undefined behaviour. You can only safely
use an object that has had const cast away as a non-const object if it
wasn't originally declared const.

Tom
--
C++ FAQ: http://www.parashift.com/c++-faq-lite/
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html

[klaus triendl]

"tom_usenet" <to********@hotmail.com> wrote in message
news:<a2********************************@4ax.com>. ..

On Wed, 02 Jun 2004 08:09:01 GMT, "klaus triendl" <tr********@mbox.at>
wrote:

hi,

recently i discovered a memory leak in our code;
I think you mean "use of a dangling reference" rather than "memory
leak".

in our code the returned object is a reference counting pointer and the
string "test" is a new object which is not freed causing a memory leak
(reported by the debugger as such).
after some investigation i

could reduce it to the following problem:
return objects of functions are handled as temporary objects, hence their
dtor is called immediately and not at the end of the function. to be able touse return objects (to avoid copying) i often assign them to a const
reference.
now, casting a const return object from a function to a non-const referenceto this return object calls immediately the dtor of the return object
anyway, any further operation deals with a non-valid object then. if i do
this in two steps - first holding a const reference to the return object andthen const_casting it, everything works like i expected it.

does anybody know whether the compiler behaves correctly?
Yes, binding a temporary directly to a const reference extends the
lifetime of the temporary to match that of the reference. You can't
bind a temporary to a non-const reference, so there's no way of doing
the lifetime extension using a non-const ref.

if the temporary is non-const i can bind it to a non-const reference. at
least with vc++7 it is possible.
Note, since you return a const string, using the returned string as a
non-const string results in undefined behaviour. You can only safely
use an object that has had const cast away as a non-const object if it
wasn't originally declared const.

well, that's an argument; and i can easily solve that problem.
but my question still remains whether the const_cast in the function
"funcWmemleak" is a good reason that the non-const reference is a non-valid
object after the assignment or not.

--
klaus triendl

[tom_usenet]

On Wed, 02 Jun 2004 14:07:20 GMT, "klaus triendl" <tr********@mbox.at>
wrote:

Yes, binding a temporary directly to a const reference extends the
lifetime of the temporary to match that of the reference. You can't
bind a temporary to a non-const reference, so there's no way of doing
the lifetime extension using a non-const ref.if the temporary is non-const i can bind it to a non-const reference. at
least with vc++7 it is possible.

This is a non-conforming compiler extension. On VC7.1 I get a warning,
and compiling with /Za (ISO mode) I get an error.

Note, since you return a const string, using the returned string as a
non-const string results in undefined behaviour. You can only safely
use an object that has had const cast away as a non-const object if it
wasn't originally declared const.well, that's an argument; and i can easily solve that problem.

Indeed, that paragraph was just an aside.
but my question still remains whether the const_cast in the function
"funcWmemleak" is a good reason that the non-const reference is a non-valid
object after the assignment or not.

If you perform the const_cast, then you are not directly binding the
reference to the temporary, and the lifetime will not be extended,
hence you have a "dangling reference" that you can't use. It is
synonymous to this conforming code that exhibits the same problem, and
doesn't use any Microsoft extensions:

#include <string>
#include <iostream>
using namespace std;

int main()
{
string const& s = static_cast<string const&>(string("foo"));
cout << s << '\n';
}

Because s isn't bound directly to the temporary (it is bound to the
result of the static cast), it doesn't extend the temporary's lifetime
and the cout call has undefined behaviour since the temporary has
already been destroyed. Remove the static_cast and it's fine. See
12.2/5 in the C++ standard.

Clear?

Tom
--
C++ FAQ: http://www.parashift.com/c++-faq-lite/
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html

[Dave Moore]

"klaus triendl" <tr********@mbox.at> wrote in message news:<sa******************@news.chello.at>...

after some investigation i

could reduce it to the following problem:
return objects of functions are handled as temporary objects, hence their
dtor is called immediately and not at the end of the function. to be able touse return objects (to avoid copying) i often assign them to a const
reference.
now, casting a const return object from a function to a non-const referenceto this return object calls immediately the dtor of the return object
anyway, any further operation deals with a non-valid object then. if i do
this in two steps - first holding a const reference to the return object andthen const_casting it, everything works like i expected it.

does anybody know whether the compiler behaves correctly?

Yes, binding a temporary directly to a const reference extends the
lifetime of the temporary to match that of the reference. You can't
bind a temporary to a non-const reference, so there's no way of doing
the lifetime extension using a non-const ref.

if the temporary is non-const i can bind it to a non-const reference. at
least with vc++7 it is possible.

Note, since you return a const string, using the returned string as a
non-const string results in undefined behaviour. You can only safely
use an object that has had const cast away as a non-const object if it
wasn't originally declared const.

well, that's an argument; and i can easily solve that problem.
but my question still remains whether the const_cast in the function
"funcWmemleak" is a good reason that the non-const reference is a non-valid
object after the assignment or not.

Yes, it is a valid reference until the const reference that the
temporary was originally bound to is destroyed (i.e. goes out of
scope), provided that the destructor for the original object is not
called in the interim (which is highly unlikely given the context).

HTH, Dave Moore

[klaus triendl]

> Because s isn't bound directly to the temporary (it is bound to the

result of the static cast), it doesn't extend the temporary's lifetime
and the cout call has undefined behaviour since the temporary has
already been destroyed. Remove the static_cast and it's fine. See
12.2/5 in the C++ standard.

Clear?

clear as glass :)
thx for the explanation. it's good to know that the compiler considers the
const_cast a function returning a result.

and thx to dave moore for your posting.
klaus