address in memory

[tom]

He
How to get address in memory?

for example:
I would like 'ADR' be the adress in memory where 'srtBuf' starts
from debuger I know it is 0x003b1460 but how to get it by program?

char* srtBuf = new char[20];
long int ADR = 0x003b1460;

t.

[Marcin Gabryszewski]

tom napisa³(a):

He
How to get address in memory?

for example:
I would like 'ADR' be the adress in memory where 'srtBuf' starts
from debuger I know it is 0x003b1460 but how to get it by program?

char* srtBuf = new char[20];
long int ADR = 0x003b1460;

t.

intptr_t ADR = reinterpret_cast<intptr_t>(srtBuf);

--
Marcin Gabryszewski
G DATA Software
www.gdata.pl

address:<FirstName><dot><Surname><at><gdata><dot>< pl>

[Michiel.Salters]

tom wrote:

for example:
I would like 'ADR' be the adress in memory where 'srtBuf' starts

char* srtBuf = new char[20];
long int ADR = 0x003b1460;

Simple:

void* ADR = srtBuf;

A void* is what you use to denote an address in memory when you don't
care about the type. You can print it using std::cout << ADR <<
std::endl;

HTH,
Michiel Salters

[Howard]

"tom" <[e-mail] in message news:ee**********@atlantis.news.tpi.pl...

He
How to get address in memory?

for example:
I would like 'ADR' be the adress in memory where 'srtBuf' starts
from debuger I know it is 0x003b1460 but how to get it by program?

char* srtBuf = new char[20];
long int ADR = 0x003b1460;

The srtBuf variable is a pointer. It holds the address of the start of the
buffer. If you want a long int variable to hold a pointer address, just
assign it, with a cast:

long int ADR = reinterpret_cast<long int>(srtBuf);

Note: This assumes that a pointer can fit in a long int on your system. On
some systems, addresses may be longer (or shorter) than a long int. (I'm
guessing that's why Marcin used intptr_t. But you used a long int, so I
showed how to use that.)

-Howard