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What a pointer really points to

P: n/a
I found the example code below, listed in the book described here:
http://cartan.cas.suffolk.edu/moin/OopDocbookWiki

The result was a bit surprising. I guess it falls into the category
of "that's what you get for lying". Can the behavior demonstrated be
explained in standardese? Yes, I know the exact result is "undefined
behavior". I can see what happened. But what was the actual violation?

// Miles are converted to kilometers.
#include <QTextStream>

QTextStream cin(stdin, QIODevice::ReadOnly);
QTextStream cout(stdout, QIODevice::WriteOnly);
QTextStream cerr(stderr, QIODevice::WriteOnly);

const double m2k = 1.609; // conversion constant

inline double mi2km(int miles) {
return (miles * m2k);
}

int main() {
int miles;
double kilometers;
cout << "Enter distance in miles: " << flush;
cin >miles ;
kilometers = mi2km(miles);
cout << "This is approximately "
<< static_cast<int>(kilometers)
<< "km."<< endl;
cout << "Without the cast, kilometers = "
<< kilometers << endl;
double* dp = const_cast<double*>(&m2k);
cout << "m2k: " << m2k << endl;
cout << "&m2k: " << &m2k << " dp: " << dp << endl;
cout << "*dp: " << *dp << endl;
*dp = 1.892; /* What are we attempting to do here?*/
cout << "Can we reach this statement? " << endl;
return 0;
}

/*OUT
Enter distance in miles: 23
This is approximately 37km.
Without the cast, kilometers = 37.007
m2k: 1.609
&m2k: 0x8049048 dp: 0x8049048
*dp: 1.609
Segmentation fault
*/
--
NOUN:1. Money or property bequeathed to another by will. 2. Something handed
down from an ancestor or a predecessor or from the past: a legacy of
religious freedom. ETYMOLOGY: MidE legacie, office of a deputy, from OF,
from ML legatia, from L legare, to depute, bequeath. www.bartleby.com/61/
Dec 9 '06 #1
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15 Replies


P: n/a
Steven T. Hatton wrote:
I found the example code below, listed in the book described here:
http://cartan.cas.suffolk.edu/moin/OopDocbookWiki
Well that just goes to show you don't it ;-)

#include <quan/out/length.hpp>

int main() {
std::cout << "Enter distance in miles: ";
quan::length::mi miles;
std::cin >miles.reference_numeric_value<quan::length::mi>() ; ;
quan::length::km kilometers = miles;
std::cout << "\nThat is approximately " << kilometers << "\n\n";

std::cout << "Can we reach this statement?\n\n";
std::cout << "Sure! ... Feel the Power of The Quan :-)\n\n";
std::cout << "-------------------------------------\n";
std::cout << "Quan documentation :\n\n";
std::cout << "<http://quan.sourceforge.net>\n\n";
std::cout << "Quan download and cvs :\n\n";
std::cout << "<http://sourceforge.net/projects/quan>\n\n";
std::cout << "-------------------------------------\n";
return 0;
}
/*
output:
Enter distance in miles: 2

That is approximately 3.21869 km

Can we reach this statement?

Sure! ... Feel the Power of The Quan :-)

-------------------------------------
Quan documentation :

<http://quan.sourceforge.net>

Quan download and cvs :

<http://sourceforge.net/projects/quan>

-------------------------------------
*/

Dec 9 '06 #2

P: n/a
"Steven T. Hatton" <ch********@germania.supwrote in message
news:c_******************************@speakeasy.ne t
I found the example code below, listed in the book described here:
http://cartan.cas.suffolk.edu/moin/OopDocbookWiki

The result was a bit surprising. I guess it falls into the category
of "that's what you get for lying". Can the behavior demonstrated be
explained in standardese? Yes, I know the exact result is "undefined
behavior". I can see what happened. But what was the actual
violation?

// Miles are converted to kilometers.
#include <QTextStream>

QTextStream cin(stdin, QIODevice::ReadOnly);
QTextStream cout(stdout, QIODevice::WriteOnly);
QTextStream cerr(stderr, QIODevice::WriteOnly);

const double m2k = 1.609; // conversion constant

inline double mi2km(int miles) {
return (miles * m2k);
}

int main() {
int miles;
double kilometers;
cout << "Enter distance in miles: " << flush;
cin >miles ;
kilometers = mi2km(miles);
cout << "This is approximately "
<< static_cast<int>(kilometers)
<< "km."<< endl;
cout << "Without the cast, kilometers = "
<< kilometers << endl;
double* dp = const_cast<double*>(&m2k);
cout << "m2k: " << m2k << endl;
cout << "&m2k: " << &m2k << " dp: " << dp << endl;
cout << "*dp: " << *dp << endl;
*dp = 1.892; /* What are we attempting to do here?*/
cout << "Can we reach this statement? " << endl;
return 0;
}

/*OUT
Enter distance in miles: 23
This is approximately 37km.
Without the cast, kilometers = 37.007
m2k: 1.609
&m2k: 0x8049048 dp: 0x8049048
*dp: 1.609
Segmentation fault
*/
Deleting the irrelevant, you have:

const double m2k = 1.609;

int main()
{
double* dp = const_cast<double*>(&m2k);
*dp = 1.892;
return 0;
}

Thus dp points to a const variable and you attempt to change that const
variable using a dereferenced dp.

The operating system has presumably stored m2k in a read-only section of
memory, so raises a segmentation fault when you attempt to write to that
memory.

--
John Carson
Dec 9 '06 #3

P: n/a
John Carson wrote:
"Steven T. Hatton" <ch********@germania.supwrote in message
news:c_******************************@speakeasy.ne t
>I found the example code below, listed in the book described here:
http://cartan.cas.suffolk.edu/moin/OopDocbookWiki

The result was a bit surprising. I guess it falls into the category
of "that's what you get for lying". Can the behavior demonstrated be
explained in standardese? Yes, I know the exact result is "undefined
behavior". I can see what happened. But what was the actual
violation?

// Miles are converted to kilometers.
#include <QTextStream>

QTextStream cin(stdin, QIODevice::ReadOnly);
QTextStream cout(stdout, QIODevice::WriteOnly);
QTextStream cerr(stderr, QIODevice::WriteOnly);

const double m2k = 1.609; // conversion constant

inline double mi2km(int miles) {
return (miles * m2k);
}

int main() {
int miles;
double kilometers;
cout << "Enter distance in miles: " << flush;
cin >miles ;
kilometers = mi2km(miles);
cout << "This is approximately "
<< static_cast<int>(kilometers)
<< "km."<< endl;
cout << "Without the cast, kilometers = "
<< kilometers << endl;
double* dp = const_cast<double*>(&m2k);
cout << "m2k: " << m2k << endl;
cout << "&m2k: " << &m2k << " dp: " << dp << endl;
cout << "*dp: " << *dp << endl;
*dp = 1.892; /* What are we attempting to do here?*/
cout << "Can we reach this statement? " << endl;
return 0;
}

/*OUT
Enter distance in miles: 23
This is approximately 37km.
Without the cast, kilometers = 37.007
m2k: 1.609
&m2k: 0x8049048 dp: 0x8049048
*dp: 1.609
Segmentation fault
*/

Deleting the irrelevant, you have:
It's often nice to have something that actually compiles and runs.
Compiling the code using the Standard Library would take a trivial amount
of effort.
const double m2k = 1.609;

int main()
{
double* dp = const_cast<double*>(&m2k);
*dp = 1.892;
return 0;
}

Thus dp points to a const variable
What does the Standard specify it should point to, or is that specified?
and you attempt to change that const
variable using a dereferenced dp.

The operating system has presumably stored m2k in a read-only section of
memory, so raises a segmentation fault when you attempt to write to that
memory.
As I sated, I understand what happened. I'm just not sure what rule was
violated. I guess I go back to slogging my way through the artful prose of
the formal arcana of the Standard.

--
NOUN:1. Money or property bequeathed to another by will. 2. Something handed
down from an ancestor or a predecessor or from the past: a legacy of
religious freedom. ETYMOLOGY: MidE legacie, office of a deputy, from OF,
from ML legatia, from L legare, to depute, bequeath. www.bartleby.com/61/
Dec 9 '06 #4

P: n/a

Steven T. Hatton wrote:
What does the Standard specify it should point to, or is that specified?
The only time you can use const_cast on some object reference is when
you *know* that the real object is not const. Any const cast on a
physically const object results in undefined behaviour. The better
option on stuff you *own* is to use mutable: That is my understanding
anyway:

struct my{
private:
int n;
public:
my():n(1),y(2){}
// mod is declared as a const function
// but maybe want to keep internal count
void mod1()const
{
// dodgy If someone declares a const my
my& mm = const_cast<my&>(*this);
++mm.n;
}
private:
mutable int y;
public:
void mod2()const
{ // always ok
++y;
}
};

int main()
{
my x;
x.mod1();
x.mod2();
}

regards
Andy Little

Dec 9 '06 #5

P: n/a
Steven T. Hatton wrote:
What does the Standard specify it should point to, or is that specified?
The only time you can use const_cast on some object reference is when
you *know* that the real object is not const. Any const cast on a
physically const object results in undefined behaviour. The better
option on stuff you *own* is to use mutable: That is my understanding
anyway:

struct my{
private:
int n;
public:
my():n(1),y(2){}
// mod is declared as a const function
// but maybe want to keep internal count
void mod1()const
{
// dodgy If someone declares a const my
my& mm = const_cast<my&>(*this);
++mm.n;
}
private:
mutable int y;
public:
void mod2()const
{ // always ok
++y;
}
};

int main()
{
my x;
x.mod1();
x.mod2();
}

regards
Andy Little

Dec 9 '06 #6

P: n/a
Steven T. Hatton wrote:
John Carson wrote:
>"Steven T. Hatton" <ch********@germania.supwrote in message
news:c_******************************@speakeasy.n et
>>I found the example code below, listed in the book described here:
http://cartan.cas.suffolk.edu/moin/OopDocbookWiki

The result was a bit surprising. I guess it falls into the category
of "that's what you get for lying". Can the behavior demonstrated be
explained in standardese? Yes, I know the exact result is "undefined
behavior". I can see what happened. But what was the actual
violation?

// Miles are converted to kilometers.
#include <QTextStream>

QTextStream cin(stdin, QIODevice::ReadOnly);
QTextStream cout(stdout, QIODevice::WriteOnly);
QTextStream cerr(stderr, QIODevice::WriteOnly);

const double m2k = 1.609; // conversion constant

inline double mi2km(int miles) {
return (miles * m2k);
}

int main() {
int miles;
double kilometers;
cout << "Enter distance in miles: " << flush;
cin >miles ;
kilometers = mi2km(miles);
cout << "This is approximately "
<< static_cast<int>(kilometers)
<< "km."<< endl;
cout << "Without the cast, kilometers = "
<< kilometers << endl;
double* dp = const_cast<double*>(&m2k);
cout << "m2k: " << m2k << endl;
cout << "&m2k: " << &m2k << " dp: " << dp << endl;
cout << "*dp: " << *dp << endl;
*dp = 1.892; /* What are we attempting to do here?*/
cout << "Can we reach this statement? " << endl;
return 0;
}

/*OUT
Enter distance in miles: 23
This is approximately 37km.
Without the cast, kilometers = 37.007
m2k: 1.609
&m2k: 0x8049048 dp: 0x8049048
*dp: 1.609
Segmentation fault
*/

Deleting the irrelevant, you have:

It's often nice to have something that actually compiles and runs.
Compiling the code using the Standard Library would take a trivial amount
of effort.
>const double m2k = 1.609;

int main()
{
double* dp = const_cast<double*>(&m2k);
*dp = 1.892;
return 0;
}

Thus dp points to a const variable

What does the Standard specify it should point to, or is that specified?
It is specified in [5.2.11/3].

>and you attempt to change that const
variable using a dereferenced dp.

The operating system has presumably stored m2k in a read-only section of
memory, so raises a segmentation fault when you attempt to write to that
memory.

As I sated, I understand what happened. I'm just not sure what rule was
violated.
The code has undefined behavior according to [7.1.5.1/4].

I guess I go back to slogging my way through the artful prose
of the formal arcana of the Standard.

Best

Kai-Uwe Bux
Dec 9 '06 #7

P: n/a
On 2006-12-09 14:14, Steven T. Hatton wrote:
John Carson wrote:
>"Steven T. Hatton" <ch********@germania.supwrote in message
news:c_******************************@speakeasy.n et
>>I found the example code below, listed in the book described here:
http://cartan.cas.suffolk.edu/moin/OopDocbookWiki

The result was a bit surprising. I guess it falls into the category
of "that's what you get for lying". Can the behavior demonstrated be
explained in standardese? Yes, I know the exact result is "undefined
behavior". I can see what happened. But what was the actual
violation?

// Miles are converted to kilometers.
#include <QTextStream>

QTextStream cin(stdin, QIODevice::ReadOnly);
QTextStream cout(stdout, QIODevice::WriteOnly);
QTextStream cerr(stderr, QIODevice::WriteOnly);

const double m2k = 1.609; // conversion constant

inline double mi2km(int miles) {
return (miles * m2k);
}

int main() {
int miles;
double kilometers;
cout << "Enter distance in miles: " << flush;
cin >miles ;
kilometers = mi2km(miles);
cout << "This is approximately "
<< static_cast<int>(kilometers)
<< "km."<< endl;
cout << "Without the cast, kilometers = "
<< kilometers << endl;
double* dp = const_cast<double*>(&m2k);
cout << "m2k: " << m2k << endl;
cout << "&m2k: " << &m2k << " dp: " << dp << endl;
cout << "*dp: " << *dp << endl;
*dp = 1.892; /* What are we attempting to do here?*/
cout << "Can we reach this statement? " << endl;
return 0;
}

/*OUT
Enter distance in miles: 23
This is approximately 37km.
Without the cast, kilometers = 37.007
m2k: 1.609
&m2k: 0x8049048 dp: 0x8049048
*dp: 1.609
Segmentation fault
*/

Deleting the irrelevant, you have:

It's often nice to have something that actually compiles and runs.
Compiling the code using the Standard Library would take a trivial amount
of effort.
>const double m2k = 1.609;

int main()
{
double* dp = const_cast<double*>(&m2k);
*dp = 1.892;
return 0;
}

Thus dp points to a const variable

What does the Standard specify it should point to, or is that specified?
It should point to the const double m2k.
>and you attempt to change that const
variable using a dereferenced dp.

The operating system has presumably stored m2k in a read-only section of
memory, so raises a segmentation fault when you attempt to write to that
memory.

As I sated, I understand what happened. I'm just not sure what rule was
violated. I guess I go back to slogging my way through the artful prose of
the formal arcana of the Standard.
The violation is trying to change the value of a const. On your system
this happens to result in a segmentation fault, but there are systems
and situations where it won't.

--
Erik Wikström
Dec 9 '06 #8

P: n/a
kwikius wrote:
Steven T. Hatton wrote:
>What does the Standard specify it should point to, or is that specified?

The only time you can use const_cast on some object reference is when
you *know* that the real object is not const. Any const cast on a
physically const object results in undefined behaviour. The better
option on stuff you *own* is to use mutable:
My inclination is to try and figure out what I'm doing wrong, and fix it.
I've come across a couple places recently where experts "Josuttis,
Vandevoorde, Blanchette, Summerfield, etc." have used const_cast<and
defended their choice. Nonetheless...
That is my understanding
anyway:

struct my{
private:
int n;
public:
my():n(1),y(2){}
// mod is declared as a const function
// but maybe want to keep internal count
void mod1()const
{
// dodgy If someone declares a const my
my& mm = const_cast<my&>(*this);
++mm.n;
}
private:
mutable int y;
public:
void mod2()const
{ // always ok
++y;
}
};

int main()
{
my x;
x.mod1();
x.mod2();
}
I know what Stroustrup had to say about mutable. He thinks it is almost
always a bad idea because it defeats the purpose of const. But the same
goes for const_cast only more so.

Here's my hack on the subsequent example from
http://cartan.cas.suffolk.edu/moin/OopDocbookWiki

#include <iostream>
using namespace std;

void el() {
const int L = 99;
int * pL = const_cast<int*>(&L);
*pL = 101;
cout << L << '\t' << &L << endl;
cout << *pL << '\t' << pL << endl;
}

void em() {
const int M = 29;
int * pM = const_cast<int*>(&M);
*pM = 66;
cout << M << '\t' << &M << endl;
cout << *pM << '\t' << pM << endl;
}

const int N = 22;
void en() {

int * pN = const_cast<int*>(&N);
*pN = 33;
cout << N << '\t' << &N << endl;
cout << *pN << '\t' << pN << endl;
}

int main() {
el();
em();
en();
}
/*
Sat Dec 09 09:06:25:c++ -o cc1 constcast1.cpp
hattons@ljosalfr:~/code/c++/ezust/src/casts/
Sat Dec 09 09:06:46:./cc1
99 0xbfc2ccb0
101 0xbfc2ccb0
29 0xbfc2ccb0
66 0xbfc2ccb0
Segmentation fault

*/

The authors are saying that[*] "const int is in stack storage class". I
believe that is old-timer talk meaning 'automatic variable' but I'm not
sure. I have a couple of observations about the code above, and the result
of executing it. The addresses of all the variables which were printed
prior to the segfault are the same. The segfault happened when the
variable was placed at global scope. If nothing else, it shows that
undefined behavior is undefined.
[*]If I understand correctly.
--
NOUN:1. Money or property bequeathed to another by will. 2. Something handed
down from an ancestor or a predecessor or from the past: a legacy of
religious freedom. ETYMOLOGY: MidE legacie, office of a deputy, from OF,
from ML legatia, from L legare, to depute, bequeath. www.bartleby.com/61/
Dec 9 '06 #9

P: n/a

Steven T. Hatton wrote:
Here's my hack on the subsequent example from
http://cartan.cas.suffolk.edu/moin/OopDocbookWiki
>
#include <iostream>
using namespace std;

void el() {
const int L = 99;
int * pL = const_cast<int*>(&L); // use pL and its undefined
*pL = 101; //undefined behaviour
cout << L << '\t' << &L << endl;
cout << *pL << '\t' << pL << endl;
}

void em() {
const int M = 29;
int * pM = const_cast<int*>(&M); // same applies
*pM = 66; //undefined behaviour
cout << M << '\t' << &M << endl;
cout << *pM << '\t' << pM << endl;
}

const int N = 22;
void en() {

int * pN = const_cast<int*>(&N); // and here
*pN = 33; //undefined behaviour
cout << N << '\t' << &N << endl;
cout << *pN << '\t' << pN << endl;
}
All undefined . The only time you can use const_cast is if you know
something is not but looks like it is:
// object pointed by p is 'logically const'
int f( const int & p)
{
int & pp = const_cast<int&>(p);
++pp; // behaviour here dependent on whether
// object referenced by pp is "physically const" or not
// if object is really const... undefined behaviour

return pp;
}

int main()
{
int x=0;
f(x); // OK... x is not really const

const int y =0;
f(y); // Bad.. y Is really const ...undefined behaviour
}

regards
Andy Little

Dec 9 '06 #10

P: n/a

Steven T. Hatton wrote:
The authors are saying that[*] "const int is in stack storage class". I
believe that is old-timer talk meaning 'automatic variable' but I'm not
sure. I have a couple of observations about the code above, and the result
of executing it. The addresses of all the variables which were printed
prior to the segfault are the same. The segfault happened when the
variable was placed at global scope. If nothing else, it shows that
undefined behavior is undefined.
Compiler writers will take any opportunity they can for optimsiation.
If they are allowed to not provide any storage for a const then they
will try not to do so if its within the rules!. The better the compiler
the more likely it is.

regards
Andy Little

Dec 9 '06 #11

P: n/a

Steven T. Hatton wrote:
I know what Stroustrup had to say about mutable. He thinks it is almost
always a bad idea because it defeats the purpose of const. But the same
goes for const_cast only more so.
They are different because mutable doesnt cause undefined behaviour,
whereas const_cast is practically asking for it.
Mutable is safe whereas const is not, but it probably means that big
opportunities for optimisation are thrown away. I have used it once
IIRC, when I should probably have changed the design.

regards
Andy Little

Dec 9 '06 #12

P: n/a
Kai-Uwe Bux wrote:
Steven T. Hatton wrote:
>John Carson wrote:
>>"Steven T. Hatton" <ch********@germania.supwrote in message
news:c_******************************@speakeasy. net
I found the example code below, listed in the book described here:
http://cartan.cas.suffolk.edu/moin/OopDocbookWiki

The result was a bit surprising. I guess it falls into the category
of "that's what you get for lying". Can the behavior demonstrated be
explained in standardese? Yes, I know the exact result is "undefined
behavior". I can see what happened. But what was the actual
violation?

// Miles are converted to kilometers.
#include <QTextStream>

QTextStream cin(stdin, QIODevice::ReadOnly);
QTextStream cout(stdout, QIODevice::WriteOnly);
QTextStream cerr(stderr, QIODevice::WriteOnly);

const double m2k = 1.609; // conversion constant

inline double mi2km(int miles) {
return (miles * m2k);
}

int main() {
int miles;
double kilometers;
cout << "Enter distance in miles: " << flush;
cin >miles ;
kilometers = mi2km(miles);
cout << "This is approximately "
<< static_cast<int>(kilometers)
<< "km."<< endl;
cout << "Without the cast, kilometers = "
<< kilometers << endl;
double* dp = const_cast<double*>(&m2k);
cout << "m2k: " << m2k << endl;
cout << "&m2k: " << &m2k << " dp: " << dp << endl;
cout << "*dp: " << *dp << endl;
*dp = 1.892; /* What are we attempting to do here?*/
cout << "Can we reach this statement? " << endl;
return 0;
}

/*OUT
Enter distance in miles: 23
This is approximately 37km.
Without the cast, kilometers = 37.007
m2k: 1.609
&m2k: 0x8049048 dp: 0x8049048
*dp: 1.609
Segmentation fault
*/

Deleting the irrelevant, you have:

It's often nice to have something that actually compiles and runs.
Compiling the code using the Standard Library would take a trivial amount
of effort.
>>const double m2k = 1.609;

int main()
{
double* dp = const_cast<double*>(&m2k);
*dp = 1.892;
return 0;
}

Thus dp points to a const variable

What does the Standard specify it should point to, or is that specified?

It is specified in [5.2.11/3].
"For two pointer types T1 and T2 where

T1 is cv1 , 0 pointer to cv1 , 1 pointer to . . . cv1 ,n ? 1 pointer to
cv1 ,n T

and

T2 is cv2 , 0 pointer to cv2 , 1 pointer to . . . cv2 ,n ? 1 pointer to
cv2 ,n T

where T is any object type or the void type and where cv1 ,k and cv2 ,k may
be different cv-qualifications, an rvalue of type T1 may be explicitly
converted to the type T2 using a const_cast. The result of a pointer
const_cast refers to the original object."

>
>>and you attempt to change that const
variable using a dereferenced dp.

The operating system has presumably stored m2k in a read-only section of
memory, so raises a segmentation fault when you attempt to write to that
memory.

As I sated, I understand what happened. I'm just not sure what rule was
violated.

The code has undefined behavior according to [7.1.5.1/4].
"Except that any class member declared mutable (7.1.1) can be modified, any
attempt to modify a const object during its lifetime (3.8) results in
undefined behavior."

So if an object is defined non-const,

int obj;

and passed through a const reference

void foo(const int& cobj){
//cobj is a const reference to a non-const object
}

the object in the body of the function to which it was passed was itself not
declared const. (cobj is a const reference to non-const obj) But the author
of a function who casts away constness has no a priori means of knowing
whether what will be passed will be const.

I guess all of the above means I can't complain about anything that happens
after *pL = 101;

void el() {
const int L = 99;
int * pL = const_cast<int*>(&L);
*pL = 101;
cout << L << '\t' << &L << endl;
cout << *pL << '\t' << pL << endl;
}
>I guess I go back to slogging my way through the artful prose
of the formal arcana of the Standard.
Thanks!
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NOUN:1. Money or property bequeathed to another by will. 2. Something handed
down from an ancestor or a predecessor or from the past: a legacy of
religious freedom. ETYMOLOGY: MidE legacie, office of a deputy, from OF,
from ML legatia, from L legare, to depute, bequeath. www.bartleby.com/61/
Dec 9 '06 #13

P: n/a
"kwikius" <an**@servocomm.freeserve.co.ukwrote in
news:11**********************@l12g2000cwl.googlegr oups.com:
>
Steven T. Hatton wrote:
>What does the Standard specify it should point to, or is that specified?

The only time you can use const_cast on some object reference is when
you *know* that the real object is not const. Any const cast on a
physically const object results in undefined behaviour. The better
option on stuff you *own* is to use mutable: That is my understanding
anyway:
Uh, not true. Casting away the constness of the physically const object is
fine. It's when one attempts to then modify the object is when Undefined
Behaviour occurs.
Dec 9 '06 #14

P: n/a
Steven T. Hatton wrote:
Kai-Uwe Bux wrote:
[snip]
>The code has undefined behavior according to [7.1.5.1/4].

"Except that any class member declared mutable (7.1.1) can be modified,
any attempt to modify a const object during its lifetime (3.8) results in
undefined behavior."

So if an object is defined non-const,

int obj;

and passed through a const reference

void foo(const int& cobj){
//cobj is a const reference to a non-const object
}

the object in the body of the function to which it was passed was itself
not declared const. (cobj is a const reference to non-const obj) But the
author of a function who casts away constness has no a priori means of
knowing whether what will be passed will be const.
It's even worse: because of 8.3.5/5 the programmer cannot even be sure that
the passed parameter object binds directly to the const reference; a
temporary copy could be created. Once you access this through const_cast,
you may just modify the temporary.

There is, as far as I can see, only one valid use of const_cast: you can use
it to define a member function returning a non-const reference / pointer to
non-const exposing a member in terms of a const member function that
exposes const access to that member. This can avoid code duplication, e.g.,
when implementing smart pointers.
[snip]
Best

Kai-Uwe Bux
Dec 9 '06 #15

P: n/a
IR
Andre Kostur wrote:
Uh, not true. Casting away the constness of the physically const
object is fine. It's when one attempts to then modify the object
is when Undefined Behaviour occurs.
Of course, you are right. But what's the point of casting constness
away if you don't modify the object? (note that I'm not including
poorly written libraries or APIs that are not const-correct)

This is the very same problem as reinterpret_cast: you can cast
anything to whatever you like as long as you cast it back to the
original type *before* trying to access the object.

IMO this is bad programming practice and should be avoided by any
means (unless you really have to cope with poorly written external
libraries).
Cheers,
--
IR
Dec 9 '06 #16

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