memory address and c for linux

darkseed2g3 · 10-06-2003, 09:12 PM

Basicly this is what i wanna do

lets say i have this code

main (int x )
{

x = 1024;

printf("%p is the memory addy " , &x);

}

of course its going to print the memory address is there any way i can change the value of it
like is there a way i can write an other program and run them both at the same time like when i run the 1st program it puts the variable x into memory with the value of x as 1024 but when i run the second one it will overide the value of x to something else basicly a buffer exploit i guess

h/w · 10-06-2003, 09:20 PM

im still getting the hang of C (as can be seen from previous posts), but the address spaces for the variables will be allocated and freed once the code block is completed. so once u ran this program, by the end of it, it would have freed up the memory (as it is a local var)?
i would use pointers and threads for this - to pass the location of the var u got here to the other code block which runs as the thread...

thats my ...

Xiangbuilder · 10-06-2003, 09:30 PM

I am not sure I understanded your meaning exactly, if so please excuse me.

Code:

#include<iostream>
using namespace std;
int main ()
{
   int x=1024;
   cout<<"x="<<x<<endl;
   cout<<"&x="<<&x<<endl;
   int* p=&x;
   cout<<"p="<<p<<endl;
   cout<<"*p="<<*p<<endl;
   *p=5;
   cout<<"p="<<p<<endl;
   cout<<"x="<<x<<endl;
}

[root@localhost linux_question]# g++ pa.cpp -o pa
[root@localhost linux_question]# ./pa
x=1024
&x=0xbfffdc94
p=0xbfffdc94
*p=1024
p=0xbfffdc94
x=5
[root@localhost linux_question]#

h/w · 10-06-2003, 09:33 PM

i thought he wanted to run 2 different processes/programs - one operating on the other? no?

Xiangbuilder · 10-06-2003, 09:38 PM

perhaps the standard c system() function can do this.

edit: sorry for my mistake.

zju_zhangxm · 10-06-2003, 10:18 PM

I think there is no way to do that.

Every process has his own stack memory, and of course they are protected by Kernel. So the other process can not "see" this process's stack.

Only kernel module( kernel program) can "see" this stack memory. But indeed it does not know exactly which is "x" value.

Kumar · 10-07-2003, 12:23 AM

No, you cannot do that. Simple reason, because both the program will have different address space and hence cannot interfere with each others address space. The address which will be printed on the screen will be the logical address and not the actual physical address. Use IPC if you want to setup communication between the two programs.

dimm_coder · 10-07-2003, 07:20 AM

Yes, as it have been mentioned above, every process has its own memory address space, so it cannot simply change address space which belongs to other process. The case for U is some kind of IPC. The best in this case is shared memory.
See man shm_open, mmap, ... for POSIX standard IPC,
man shmget, ... for UNIX98 IPC.

orgcandman · 10-07-2003, 11:52 AM

You're kindof right....I'm going to take a stab at explaining this in detail (please excuse errors)

In the world of execution, there are two major modes of operation under the x86 architecture: Real mode and Protected Mode. They each deal with the way memory becomes segmented for the execution of the program. In the case of protected mode (the "right" mode as far as OS developers are concerned), when a piece of code tries to access new memory, it throws the appropriate Page Fault exception which causes the kernel to look at the process and handle the exception properly. IE: a malloc() will throw a request for new pages page fault exception and the kernel will either return a new block of memory, or a NULL if no memory is available.
In protected, a piece of code entering another process' address space throws a Segmentation Page Fault exception. This is why that process dies with a SYSSEGV error code. Under this system, what he's trying to accomplish will never happen.

Under real mode, however, the system doesn't use a paged approach to memory (which is why there is a significantly smaller amount of memory available to the process running) and a program can fairly easily encraoch on another's address space (corewars without an emulator

). This is why DOS had some crazy issues with multi-tasking programs. Sometimes, even, a process under real mode would accidentally overwrite the kernel routines and cause the OS to do funky things. Under that system declaring a pointer to memory like char *p = (char *)<address> would surely get you access to any arbitrary point in memory.

Even under protected mode, if you're executing in kernel space, you can access raw points of memory (how do you think the video driver works?? char *vid = (char *)0xB80000; for vga and char *vid = 0x080000; for monochrome

) but doing such things is risky and should be avoided at all costs.

This is, I believe, the information relevant for the x86 platform. This all translates into, under a proper OS, you can't accomplish this.

Aaron

jinksys · 10-07-2003, 04:30 PM

nice explanation

darkseed2g3 · 10-09-2003, 09:21 PM

basiclly, i was talking about a buffer overrun
when you have a buffer defined for a program i was trying to overrun the buffer and have it point to another instruction

i was just wondering how to do it easier or if anyone has anything that would explian it better . I have read phrack 49 - smashing the stack but i need more.

you can all understand that right.

jinksys · 10-09-2003, 09:50 PM

Well, you probebly not going to understand smashing the stack unless you are familiar with assembly language.

Do you want me to post a sample program that executes a buffer overrun?

Xiangbuilder · 10-09-2003, 10:57 PM

"Do you want me to post a sample program that executes a buffer overrun?"
I want to learn from you.
Thank you.

h/w · 10-10-2003, 03:25 PM

i would like to see some code that does it too, so if you could post it, it owuld be great. umm, it wouldnt be some code that does a gets() in a loop. would it?

jinksys · 10-10-2003, 03:57 PM

Ill post a simple buffer overrun program when i get home,
about 5:15 CDT