Can I modify a constant variable through a pointer obtained from const_cast?

[CoolPint]

While I was reading about const_cast, I got curious and wanted to know
if I could modify a constant variable through a pointer which has been
"const_cast "ed. Since the pointer would be pointing to the constant
variable and if I changed the value through the pointer, the constant
variable should have been modified. Well, that's what I thought until
I wrote the following and run it.

#include <iostream>
using std::cout;
using std::endl;
int main ()
{
const int i = 0;
int &j = const_cast<int& >(i);
j = 10; // I thought this would change i, but it didn't
cout << &i << endl; // &i and &j are same as I expected
cout << &j << endl;
cout << i << endl; // but i is still 0
cout << j << endl; // and j is 10!

int * p = const_cast<int *>(&i);
*p = 30; // I change i via p
cout << &i << endl; // I confirm that p is pointing to i!
cout << p << endl; // I confirm that p is pointing to i
cout << i << endl; // i is still 0!
cout << *p << endl; // how can *p be 30 then?

return 0;
}

I am very much confused. &i and &j are same, meaning they refer to the
same location. p is also pointing to the same location. But then how
can i still be 0 and j be 10 and i still be 0 and *p 30????

Is j allocated new memory space? Is p pointing to a new location?

Can anyone kindly help me figure out what is going on?

BTW, is what I am observing is standard behaviour or particular to my
compiler (g++ version 3.2.3)?

Thank you very much in advance!

[Victor Bazarov]

CoolPint wrote:

While I was reading about const_cast, I got curious and wanted to know
if I could modify a constant variable through a pointer which has been
"const_cast "ed.
Let me say this right away: you can modify it, but it leads to undefined
behaviour. Once we established that, all other things are speculation,
at best.
Since the pointer would be pointing to the constant
variable and if I changed the value through the pointer, the constant
variable should have been modified. Well, that's what I thought until
I wrote the following and run it.

#include <iostream>
using std::cout;
using std::endl;
int main ()
{
const int i = 0;
Declaring a const int object does not necessarily allocate any memory
for it. Just so we're clear on that. The compiler is free to use
the actual value of the const given to it at the moment of declaration
instead of the object in memory (which, incidentally, may or may not
exist).
int &j = const_cast<int& >(i);
So far, so good.
j = 10; // I thought this would change i, but it didn't
It did. Or it didn't. Or it did, but then changed it back unbeknownst
to you. Anything can happen. No guarantees whatsoever. You broke your
own promise to the compiler: 'i' is constant. You attempted to change
it. The compiler (and the code produced by it) is not obligated to do
anything predictable or meaningful, for that matter.
cout << &i << endl; // &i and &j are same as I expected
cout << &j << endl;
cout << i << endl; // but i is still 0
Well, this is where the speculation begins. 'i' is not really 0 "still".
It has been changed. The _code_ however was created to simply use '0'
instead of 'i' for the statement above. The compiler is allowed to do
that. It optimises away memory access since you promised that the memory
is not going to be altered in any way.
cout << j << endl; // and j is 10!

int * p = const_cast<int *>(&i);
*p = 30; // I change i via p
Another cause for undefined behaviour, right here. If all bets were off
before, now they are really REALLY off.
cout << &i << endl; // I confirm that p is pointing to i!
cout << p << endl; // I confirm that p is pointing to i
cout << i << endl; // i is still 0!
cout << *p << endl; // how can *p be 30 then?

return 0;
}

I am very much confused. &i and &j are same, meaning they refer to the
same location. p is also pointing to the same location. But then how
can i still be 0 and j be 10 and i still be 0 and *p 30????

Is j allocated new memory space? Is p pointing to a new location?

Can anyone kindly help me figure out what is going on?
I hope I have. Ask more questions if you happen to have them.

BTW, is what I am observing is standard behaviour or particular to my
compiler (g++ version 3.2.3)?

Yes. And no. There is no standard behaviour. The Standard says that
the behaviour is _undefined_.

[Siemel Naran]

"CoolPint" <co******@yahoo .co.uk> wrote in message

While I was reading about const_cast, I got curious and wanted to know
if I could modify a constant variable through a pointer which has been
"const_cast "ed. Since the pointer would be pointing to the constant
variable and if I changed the value through the pointer, the constant
variable should have been modified. Well, that's what I thought until
I wrote the following and run it.
const_cast is only garaunteed to work when the original object was const.
If the original object is const, the compiler can substitute its values into
cout statements and so forth, so that even if you modify the variable
through a pointer the cout statement will print the original values. In
more complex examples, this replacing and evaluating of const variables at
compile time (I think it's called const folding) leads to great
optimizations. The compiler can also store the variable in read-only memory
as part of the compiled program, so that if you modify the variable through
a pointer the program will crash. In other cases, const_cast may work as
you want.
{
const int i = 0;
int &j = const_cast<int& >(i);
j = 10; // I thought this would change i, but it didn't
cout << &i << endl; // &i and &j are same as I expected
cout << &j << endl;
cout << i << endl; // but i is still 0
cout << j << endl; // and j is 10!

For cout << i, the compiler has substituted 0 into the cout statement as it
was originally const.

[Siemel Naran]

"Victor Bazarov" <v.********@com Acast.net> wrote in message news:O5RLc.47

#include <iostream>
using std::cout;
using std::endl;
int main ()
{
const int i = 0;

Declaring a const int object does not necessarily allocate any memory
for it. Just so we're clear on that. The compiler is free to use
the actual value of the const given to it at the moment of declaration
instead of the object in memory (which, incidentally, may or may not
exist).

True, but if you take the address of the variable and use this address, the
compiler must allocate memory for it.

int &j = const_cast<int& >(i); cout << &j << endl;

[Andre Kostur]

co******@yahoo. co.uk (CoolPint) wrote in
news:15******** *************** ***@posting.goo gle.com:

While I was reading about const_cast, I got curious and wanted to know
if I could modify a constant variable through a pointer which has been
"const_cast "ed. Since the pointer would be pointing to the constant
variable and if I changed the value through the pointer, the constant
variable should have been modified. Well, that's what I thought until
I wrote the following and run it.
No. It is your responsibility to ensure that the object that you
const_cast away from really isn't a const object. Any attempt to modify
a const object, through whatever means you can find, is undefined
behaviour.

Some in-line comments which will make more sense with the explanation
below:
#include <iostream>
using std::cout;
using std::endl;
int main ()
{
const int i = 0;
int &j = const_cast<int& >(i);
This refers to the variable i.
j = 10; // I thought this would change i, but it didn't
cout << &i << endl; // &i and &j are same as I expected
This has the address of the variable i.
cout << &j << endl;
cout << i << endl; // but i is still 0
Since i is declared as const, the compiler can remove the reference to i,
and rewrite this line to effectively be:

cout << 0 << endl;
cout << j << endl; // and j is 10!
This refers to the variable i.
int * p = const_cast<int *>(&i);
*p = 30; // I change i via p
cout << &i << endl; // I confirm that p is pointing to i!
cout << p << endl; // I confirm that p is pointing to i
cout << i << endl; // i is still 0!
cout << *p << endl; // how can *p be 30 then?
Same logic as above.

return 0;
}

I am very much confused. &i and &j are same, meaning they refer to the
same location. p is also pointing to the same location. But then how
can i still be 0 and j be 10 and i still be 0 and *p 30????
Since you told the compiler that i is a const int 0, anyplace that i is
used, the compiler could optimize the variable access away and replace it
with a literal 0. When you took the address of i, it needs a memory
location, so the compiler is forced to allocate memory for it. (Side
question: if you never cause the address of i to be taken, even
implicitly, is the compiler required to reserve memory space for i, or
can it completely optimize the variable out of existance?)
Is j allocated new memory space? Is p pointing to a new location?

Can anyone kindly help me figure out what is going on?

BTW, is what I am observing is standard behaviour or particular to my
compiler (g++ version 3.2.3)?

It's undefined behaviour. The compiler could format your hard drive and
cause monkeys to fly out of your nether regions (OK, from a Standard
point of view it could...).

[Victor Bazarov]

Siemel Naran wrote:

"Victor Bazarov" <v.********@com Acast.net> wrote in message news:O5RLc.47

#include <iostream>
using std::cout;
using std::endl;
int main ()
{
const int i = 0;

Declaring a const int object does not necessarily allocate any memory
for it. Just so we're clear on that. The compiler is free to use
the actual value of the const given to it at the moment of declaration
instead of the object in memory (which, incidentally, may or may not
exist).

True, but if you take the address of the variable and use this address, the
compiler must allocate memory for it.

Yes, but only for the purpose of taking its address. The compiler is
still free to use the value in any other instances without having to
go to the [allocated due to address taken] memory object.

V

[JKop]

Siemel Naran posted:

const_cast is only garaunteed to work when the original object was const.

Could you please elborate on that?
-JKop

[JKop]

Can some-one please explain to me why "const_cast " exists
at all? From what I can see it just produces undefined
behaviour.
-JKop

[Andre Kostur]

JKop <NU**@NULL.NULL > wrote in news:V9******** *********@news. indigo.ie:

Can some-one please explain to me why "const_cast " exists
at all? From what I can see it just produces undefined
behaviour.

You may be attempting to call legacy code which was not const-correct.
Let's assume that you have an old C library that has a function:

void fn(data * pdata);

And it is well-documented that fn() does not modify the data.

Now, in your C++ code you have:

void cfn(const data & somedata)
{
fn(&somedata);
}
OK, you _tried_ to have that, but the C++ compiler rejects it since you
are attempting to pass a const data* where the function signature is only
data*. However, this is legal and safe:

void cfn(const data & somedata)
{
fn(const_cast<d ata *>(&somedata)) ;
}

Since fn is documented to not change the passed in data, this is safe and
legal.
I suppose a more relevent example would be:

int strcmp(char * cp1, char * cp2);

And try passing in:

std::string str1("abc"), str2("def");

std::cout << strcmp(str1.c_s tr(), str2.c_str());

(assume appropriate #include's)

[Ali Cehreli]

On Thu, 22 Jul 2004 09:43:09 -0700, JKop wrote:

Siemel Naran posted:

const_cast is only garaunteed to work when the original object was

const.

I don't think this is correct.

Could you please elborate on that?

I suspect that Siemel Naran meant this: Taking away const by a
const_cast is guaranteed to work only if the original object was
non-const to begin with.

It is undefined behavior to modify the original object if it was
initially defined as const.

Ali