regarding const modification

[Vijay]

Hi ,

Code:
#include "iostream.h"

void main(void)
{
const int i=10;
int *p;
p=const_cast<int*>(&i);
*p=111;
cout<<i<<" "<<*p;
}

I am compiling on VC++ 6, and output is : 10 111

I am not getting.. why value of i and *p is different??
Can any one help me?

Thanks
Vijay

[Oregu]

1) example should be correct enough to compile and link
2) anyway: see the code disassembled - compiler replaces i by pushing
10 on stack.
Vijay wrote:

Hi ,

Code:
#include "iostream.h"

void main(void)
{
const int i=10;
int *p;
p=const_cast<int*>(&i);
*p=111;
cout<<i<<" "<<*p;
}

I am compiling on VC++ 6, and output is : 10 111

I am not getting.. why value of i and *p is different??
Can any one help me?

Thanks
Vijay

[Kai-Uwe Bux]

Vijay wrote:

Hi ,

Code:
#include "iostream.h"

void main(void)
{
const int i=10;
int *p;
p=const_cast<int*>(&i);
*p=111;
cout<<i<<" "<<*p;
}

I am compiling on VC++ 6, and output is : 10 111

I am not getting.. why value of i and *p is different??
Can any one help me?

Your mistake is to have expectations about this code, which exhibits
undefined behavior.
Best

Kai-Uwe Bux

[Sharad Kala]

"Vijay" <mt******@gmail.com> wrote in message

| Hi ,
|
| Code:
| #include "iostream.h"

Non standard header. Read about extensionless headers.

| void main(void)

Undefined behavior. It's int main()

| {
| const int i=10;
| int *p;
| p=const_cast<int*>(&i);

Undefined behavior. You can't cast away the constness of an object that was
originally const.

| *p=111;
| cout<<i<<" "<<*p;
| }

Sharad

[Vijay]

what is the wrong in this code? can u pl explain?

As I know. pointer p have the same addresh as variable i.
Then it should give same value for *p and i.

Kai-Uwe Bux wrote:

Vijay wrote:

Hi ,

Code:
#include "iostream.h"

void main(void)
{
const int i=10;
int *p;
p=const_cast<int*>(&i);
*p=111;
cout<<i<<" "<<*p;
}

I am compiling on VC++ 6, and output is : 10 111

I am not getting.. why value of i and *p is different??
Can any one help me?

Your mistake is to have expectations about this code, which exhibits
undefined behavior.
Best

Kai-Uwe Bux

[Vijay]

ok.

Can u tell me.. how can i change the value of const using pointer??
Sharad Kala wrote:

"Vijay" <mt******@gmail.com> wrote in message

| Hi ,
|
| Code:
| #include "iostream.h"

Non standard header. Read about extensionless headers.

| void main(void)

Undefined behavior. It's int main()

| {
| const int i=10;
| int *p;
| p=const_cast<int*>(&i);

Undefined behavior. You can't cast away the constness of an object that was
originally const.

| *p=111;
| cout<<i<<" "<<*p;
| }

Sharad

[Rolf Magnus]

Vijay wrote:

#include "iostream.h"

void main(void)
{
const int i=10;
int *p;
p=const_cast<int*>(&i);
*p=111;
cout<<i<<" "<<*p;
}

what is the wrong in this code? can u pl explain?
I have no idea who "u pl" is, but Sharad Kala explained it.
As I know. pointer p have the same addresh as variable i.
Then it should give same value for *p and i.

"const" means that the object is constant, i.e. must never be modified. You
used a dirty hack that works around C++'s way of ensuring that this doesn't
happen. After that, you are responsible for never changing the value of
that constant, and you failed. However, the compiler will still generate
its code under the assumption that i is never changed, because you wrote
that it is constant.

[benben]

Vijay wrote:

Hi ,

Code:
#include "iostream.h"

void main(void)
{
const int i=10;
int *p;
p=const_cast<int*>(&i);
*p=111;
cout<<i<<" "<<*p;
}

I am compiling on VC++ 6, and output is : 10 111

I am not getting.. why value of i and *p is different??
Can any one help me?
Because the program does not read from &i to get the value of i. I am
not sure whether compilers are meant to (or allowed to) do it (very
likely though) but your program is ill formed.

I would appreciate if anyone can quote the standard relevant to the topic.

Thanks
Vijay

Regards,
Ben

Regards,
Ben

[Rolf Magnus]

Vijay wrote:

ok.

Can u tell me.. how can i change the value of const using pointer??

Why do you want to? The sole purpose of const is to make the object not
modifyable. If you want to modify it, simply don't make it const.

[Sharad Kala]

"Vijay" <mt******@gmail.com> wrote in message

| ok.
|
| Can u tell me.. how can i change the value of const using pointer??

0. Please don't top post. If you don't understand what it means then search
for netiquettes on google.
1. Think what does a const imply. Const would be rendered useless if you
could get away with it using hacks.

Sharad

[snip]

| > | const int i=10;
| > | int *p;
| > | p=const_cast<int*>(&i);
| >
| > Undefined behavior. You can't cast away the constness of an object that
was
| > originally const.
| >
| > | *p=111;
| > | cout<<i<<" "<<*p;
| > | }
| >
| > Sharad
|

[Vijay]

Rolf Magnus wrote:

Vijay wrote:

#include "iostream.h"

void main(void)
{
const int i=10;
int *p;
p=const_cast<int*>(&i);
*p=111;
cout<<i<<" "<<*p;
}

what is the wrong in this code? can u pl explain?

I have no idea who "u pl" is, but Sharad Kala explained it.

As I know. pointer p have the same addresh as variable i.
Then it should give same value for *p and i.

"const" means that the object is constant, i.e. must never be modified. You
used a dirty hack that works around C++'s way of ensuring that this doesn't
happen. After that, you are responsible for never changing the value of
that constant, and you failed. However, the compiler will still generate
its code under the assumption that i is never changed, because you wrote
that it is constant.

ok. thanks for kind response.

It means... I can not change the value of const object even using
pointers.
right?

[Vijay]

what is the wrong in this code? can u pl explain?

As I know. pointer p have the same addresh as variable i.
Then it should give same value for *p and i.

Kai-Uwe Bux wrote:

Vijay wrote:

Hi ,

Code:
#include "iostream.h"

void main(void)
{
const int i=10;
int *p;
p=const_cast<int*>(&i);
*p=111;
cout<<i<<" "<<*p;
}

I am compiling on VC++ 6, and output is : 10 111

I am not getting.. why value of i and *p is different??
Can any one help me?

Your mistake is to have expectations about this code, which exhibits
undefined behavior.
Best

Kai-Uwe Bux

[Rolf Magnus]

Vijay wrote:

"const" means that the object is constant, i.e. must never be modified.
You used a dirty hack that works around C++'s way of ensuring that this
doesn't happen. After that, you are responsible for never changing the
value of that constant, and you failed. However, the compiler will still
generate its code under the assumption that i is never changed, because
you wrote that it is constant.

ok. thanks for kind response.

It means... I can not change the value of const object even using
pointers.
right?

Yup, that's right.

[Kai-Uwe Bux]

Vijay wrote:

what is the wrong in this code? can u pl explain?
Who?

As I know. pointer p have the same addresh as variable i.
Then it should give same value for *p and i.

Please do not top-post in this news group. It is considered poor style.

> void main(void)
> {
> const int i=10;
> int *p;
> p=const_cast<int*>(&i);
> *p=111;
> cout<<i<<" "<<*p;
> }

You are trying to modify a const object. This is undefined according to the
standard [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.
Asside from the standard, one can make reasonable guesses as to what your
compiler did: you promissed i to be const. Thus, whenever the compiler sees
i in your code, it is entitled to assume that the value is 10. This is
called constant propagation: any value known at compile time is precomputed
to minimize resource use at run-time. To allow for this kind of
optimization is a rational for the provision of the standard.
Best

Kai-Uwe Bux

[Vijay]

ok. I got.

Thanks a lot to all of you.

Regards,
Vijay

Rolf Magnus wrote:

Vijay wrote:

"const" means that the object is constant, i.e. must never be modified.
You used a dirty hack that works around C++'s way of ensuring that this
doesn't happen. After that, you are responsible for never changing the
value of that constant, and you failed. However, the compiler will still
generate its code under the assumption that i is never changed, because
you wrote that it is constant.

ok. thanks for kind response.

It means... I can not change the value of const object even using
pointers.
right?

Yup, that's right.

[Mike Smith]

Vijay wrote:

It means... I can not change the value of const object even using
pointers.
right?

Well, you *can* modify a const object using pointer tricks. But it is
undefined behavior according to the Standard, and you *should* not do it
unless you have a very good reason and are willing to accept
responsibility for the result.

--
Mike Smith

[Andre Kostur]

On Wed, 31 May 2006 15:33:37 +0530, Sharad Kala wrote:

"Vijay" <mt******@gmail.com> wrote in message

| {
| const int i=10;
| int *p;
| p=const_cast<int*>(&i);

Undefined behavior. You can't cast away the constness of an object that
was originally const.
Not true. You can cast away the constness all you want....

| *p=111;

....But here's the Undefined Behaviour. You're not allowed to modify that
object. const_cast is good for interfacing with older C libraries, and
other libraries which aren't const-correct. As in, they declare functions
that don't specify const, but really do treat their parameters as const.