read memory address

[Someonekicked]

Is it possible to read a memory address with C++;

For example, If I run this code first:

*************
#include <iostream>
using namespace std;
void main()
{
int *zz = new int;
*zz = 79;
cout << zz << endl;
}
*************

And suppose the output will be 00322E70 .

Now, can I write a new program in C++, and Access the memory address
00322E70 and check what is in there?

So I am thinking of something Like the following code, though the following
DOES NOT WORK:

*************
#include <iostream>
using namespace std;
void main()
{
int *zz;
zz = 00322E70;
cout << *zz << endl;
}
*************

[Victor Bazarov]

Someonekicked wrote:

Is it possible to read a memory address with C++;
Not portably, but yes.
For example, If I run this code first:

*************
#include <iostream>
using namespace std;
void main()
int main()
{
int *zz = new int;
*zz = 79;
cout << zz << endl;
}
*************

And suppose the output will be 00322E70 .

Now, can I write a new program in C++, and Access the memory address
00322E70 and check what is in there?
It's in the "virtual memory space" of your program. What physical area
it corresponds to in your platform is beyond the scope of the language and
has to be asked about in a newsgroup for your platform.
So I am thinking of something Like the following code, though the following
DOES NOT WORK:

*************
#include <iostream>
using namespace std;
void main()
int main()
{
int *zz;
zz = 00322E70;
You need to (a) form the literal correctly by prefixing it with '0x' and
(b) use reinterpret_cast<int*>:

zz = reinterpret_cast<int*>(0x00322E70);

However, if that address does not really represent an 'int', using the
pointer resulting from such conversion has undefined behaviour. So, C++
imposes no requirements on that program to do anything in particular.
cout << *zz << endl;
}
*************

V

[Phlip]

Someonekicked wrote:

For example, If I run this code first:

*************
#include <iostream>
using namespace std;
void main()
Use Google on this newsgroup for "void main", and for "using namespace std".

From whom are you learning these things?
{
int *zz = new int;
*zz = 79;
cout << zz << endl;
}
*************

And suppose the output will be 00322E70 .

Now, can I write a new program in C++, and Access the memory address
00322E70 and check what is in there?

Google for "virtual memory". Each time you run a program, a modern OS will
construct virtual memory for the program, using a secret hardware table of
offsets to real memory. Each time you run the memory could be different.

And all of this is platform specific: The C++ itself won't define the result
of pointing to a memory address that does not contain a well-formed C++
object.

--
Phlip
http://www.c2.com/cgi/wiki?ZeekLand

[ben]

you can use reinterpret_cast to achieve it but i am looking forward to some
memory violation crashes :)

ben

[Someonekicked]

thx for your reply
"Victor Bazarov" <v.********@comAcast.net> wrote in message
news:Cw****************@newsread1.mlpsca01.us.to.v erio.net...

Someonekicked wrote:

Is it possible to read a memory address with C++;

Not portably, but yes.

For example, If I run this code first:

*************
#include <iostream>
using namespace std;
void main()

int main()

{
int *zz = new int;
*zz = 79;
cout << zz << endl;
}
*************

And suppose the output will be 00322E70 .

Now, can I write a new program in C++, and Access the memory address
00322E70 and check what is in there?

It's in the "virtual memory space" of your program. What physical area
it corresponds to in your platform is beyond the scope of the language and
has to be asked about in a newsgroup for your platform.

So I am thinking of something Like the following code, though the
following DOES NOT WORK:

*************
#include <iostream>
using namespace std;
void main()

int main()

{
int *zz;
zz = 00322E70;

You need to (a) form the literal correctly by prefixing it with '0x' and
(b) use reinterpret_cast<int*>:

zz = reinterpret_cast<int*>(0x00322E70);

However, if that address does not really represent an 'int', using the
pointer resulting from such conversion has undefined behaviour. So, C++
imposes no requirements on that program to do anything in particular.

cout << *zz << endl;
}
*************

V