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Dimitar
Guest
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 Posted: Tue Apr 26, 2005 4:45 pm Post subject: Pointer to temporary |
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Why the following code (anotherFunc) is wrong:
class A
{
....
};
A func()
{
A a;
...
return a;
}
void anotherFunc()
{
const A* = &func();
// use A
}
If reference is used instead of a pointer, everything is Ok.
Probably I am missing something from the standard.
The only explanation I have is that if the temporary is not referenced it
goes out of scope and gets deleted. If pointer is used, it is much more
difficult (or even impossible) for the compiler to determine when the
temporary goes out of scope. Does this sound correct?
Any comments will be appreciated.
Thanks,
Dimitar
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Victor Bazarov
Guest
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| Posted: Wed Apr 27, 2005 8:46 am Post subject: Re: Pointer to temporary |
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Dimitar wrote:
| Quote: | Why the following code (anotherFunc) is wrong:
class A
{
...
};
A func()
{
A a;
...
return a;
}
void anotherFunc()
{
const A* = &func();
// use A
}
If reference is used instead of a pointer, everything is Ok.
Probably I am missing something from the standard.
|
Probably.
| Quote: | The only explanation I have is that if the temporary is not referenced it
goes out of scope and gets deleted. If pointer is used, it is much more
difficult (or even impossible) for the compiler to determine when the
temporary goes out of scope. Does this sound correct?
|
Yes, with the only correction. Temporaries do not "go out of scope".
Their lifetimes end.
References need to be initialised and can never be "re-seated". Pointers
can be assigned to thus making them point to a different object. In order
to keep track of the lifetime of the temporary bound to a pointer, the
compiler would have to jump through hoops figuring out whether you have
changed the pointer value. IOW, the lifetime of the pointer has no
bearing on the lifetime of the temporary if it were allowed to connect
them.
V
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Allan W
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| Posted: Wed Apr 27, 2005 8:48 am Post subject: Re: Pointer to temporary |
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Dimitar wrote (altered for brevity)
| Quote: | class A { /* ... */ };
A func() {
A a;
return a;
}
So func() returns an A by value. |
| Quote: | void anotherFunc()
{
const A* = &func();
|
This is a syntax error. I think you meant
const A*a = &func();
The return value of func() is a temporary of type A.
You take the address of that temporary and save it in a.
Then the temporary goes out of scope and gets destroyed.
Your pointer is now "dangling".
| Quote: | If reference is used instead of a pointer, everything is Ok.
|
Yes. There's an exception to the rule about the lifetime of
temporaries;
if you take a reference to it, then the lifetime extends to the
lifetime
of the reference.
| Quote: | Probably I am missing something from the standard.
|
Yes.
| Quote: | The only explanation I have is that if the temporary is not
referenced it
goes out of scope and gets deleted. If pointer is used, it is much
more
difficult (or even impossible) for the compiler to determine when the
temporary goes out of scope. Does this sound correct?
|
Pretty close.
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Ron Natalie
Guest
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| Posted: Wed Apr 27, 2005 8:48 am Post subject: Re: Pointer to temporary |
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Dimitar wrote:
| Quote: | const A* = &func();
// use A
|
| Quote: |
If reference is used instead of a pointer, everything is Ok.
|
Well the first problem is you have no identifier in the above declaration.
Yes, you are missing the fact that the temporary (the return of func)
ceases to exist at the end of the full expression it is created in.
In other words, by the time you get beyond "// use A" above it has
ceased to be.
However, there's a special feature for references that extends the
lifetime of any object bound to it until the reference itself ends.
const A& foo = func();
the returned value will stay valid as long as foo is.
If you want you can do
const A& foo = func(); // hold foo alive.
const A* fooptr = &foo; // and use foo as a pointer.
Pointers don't extend the life of temporaries because unlike references
they can be changed to refer elsewhere before their lifetime ends.
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Bronek Kozicki
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| Posted: Fri Apr 29, 2005 1:36 am Post subject: Re: Pointer to temporary |
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Allan W wrote:
| Quote: | Dimitar wrote (altered for brevity)
class A { /* ... */ };
void anotherFunc()
{
const A* = &func();
This is a syntax error. I think you meant
const A*a = &func();
|
this is still error and should not compile. Built-in operator& does not
accept temporary values. This limitation does not apply to overloaded
operator& , thus it will compile if there is one in class A . But then,
a is dangling pointer as you (and others) noted in this thread.
B.
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Dimitar
Guest
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| Posted: Sun May 01, 2005 2:59 pm Post subject: Re: Pointer to temporary |
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Well,
it compiles without problems with gcc and microsoft C++ (.net).
gcc gives warning (even if you do not specify any flags), and the ms
compiler gives a warning only if you compile with warning level W4.
Probably the next question would be: Why does it compile?
Such code does not make sense.
Dimitar
Bronek Kozicki wrote:
| Quote: | Allan W wrote:
this is still error and should not compile. Built-in operator& does not
accept temporary values. This limitation does not apply to overloaded
operator& , thus it will compile if there is one in class A . But then,
a is dangling pointer as you (and others) noted in this thread.
|
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Bronek Kozicki
Guest
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| Posted: Mon May 02, 2005 8:18 am Post subject: Re: Pointer to temporary |
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Dimitar wrote:
| Quote: | it compiles without problems with gcc and microsoft C++ (.net).
gcc gives warning (even if you do not specify any flags), and the ms
compiler gives a warning only if you compile with warning level W4.
|
Some compilers quite liberally implement rules of the C++ standard,
unless thay are told to be "strict". In MSVC use option /Za , in g++ use
-pedantic -ansi . But even with these flag g++ seems to ignore semantics
of builtin operator& .
B.
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kanze@gabi-soft.fr
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| Posted: Tue May 03, 2005 8:21 am Post subject: Re: Pointer to temporary |
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Bronek Kozicki wrote:
| Quote: | Dimitar wrote:
it compiles without problems with gcc and microsoft C++
(.net). gcc gives warning (even if you do not specify any
flags), and the ms compiler gives a warning only if you
compile with warning level W4.
Some compilers quite liberally implement rules of the C++
standard, unless thay are told to be "strict". In MSVC use
option /Za , in g++ use -pedantic -ansi . But even with these
flag g++ seems to ignore semantics of builtin operator& .
|
Are you sure. I get a hard error from g++ when taking the
address of the return value of a function, both with a
relatively recent version (3.4.3) and with a very old one
(2.95.2). Presuming, of course, that the function returns a
value; if it returns a reference, everything compiles fine, as
it should.
--
James Kanze GABI Software
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34
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Dimitar
Guest
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| Posted: Wed May 04, 2005 8:11 am Post subject: Re: Pointer to temporary |
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Below is a small test program that I compiled and run. No errors, only
single warning.
Dimitar
bash-2.05b$ more Test.cpp
#include <iostream>
using namespace std;
class A
{
public:
A()
{
cout << "CONSTRUCTOR" << endl;
}
~A()
{
cout << "DESTRUCTOR" << endl;
}
};
A ff()
{
A a;
return a;
}
int main ()
{
const A& r = ff();
cout << "Point 1" << endl;
const A* p = &ff();
cout << "Point 2" << endl;
return 0;
}
bash-2.05b$ g++ -o Test Test.cpp
Test.cpp: In function `int main()':
Test.cpp:29: warning: taking address of temporary
bash-2.05b$ Test
CONSTRUCTOR
Point 1
CONSTRUCTOR
DESTRUCTOR
Point 2
DESTRUCTOR
bash-2.05b$ g++ --version
g++ (GCC) 3.3.5 (Gentoo Linux 3.3.5-r1, ssp-3.3.2-3, pie-8.7.7.1)
Copyright (C) 2003 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
[email]kanze (AT) gabi-soft (DOT) fr[/email] wrote:
| Quote: | Bronek Kozicki wrote:
Are you sure. I get a hard error from g++ when taking the
address of the return value of a function, both with a
relatively recent version (3.4.3) and with a very old one
(2.95.2). Presuming, of course, that the function returns a
value; if it returns a reference, everything compiles fine, as
it should.
|
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Ben Hutchings
Guest
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| Posted: Thu May 05, 2005 12:16 am Post subject: Re: Pointer to temporary |
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[email]kanze (AT) gabi-soft (DOT) fr[/email] wrote:
| Quote: | Bronek Kozicki wrote:
Dimitar wrote:
it compiles without problems with gcc and microsoft C++
(.net). gcc gives warning (even if you do not specify any
flags), and the ms compiler gives a warning only if you
compile with warning level W4.
Some compilers quite liberally implement rules of the C++
standard, unless thay are told to be "strict". In MSVC use
option /Za , in g++ use -pedantic -ansi . But even with these
flag g++ seems to ignore semantics of builtin operator& .
Are you sure. I get a hard error from g++ when taking the
address of the return value of a function, both with a
relatively recent version (3.4.3) and with a very old one
(2.95.2). Presuming, of course, that the function returns a
value; if it returns a reference, everything compiles fine, as
it should.
|
The diagnostic varies depending on whether the type is a built-in or
user-defined type (reported as bug 21386).
Ben.
--
Ben Hutchings
Having problems with C++ templates? Your questions may be answered by
<http://womble.decadentplace.org.uk/c++/template-faq.html>.
[ See http://www.gotw.ca/resources/clcm.htm for info about ]
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