Thread: Segmentation fault while writing value at given address.

Hi All,

I am trying to write a value at given address location, But i am getting segmentation fault, Following is code, Please have look and help to find out the error cause.

Code:

#include <stdio.h> 


int main(){


  *(volatile int*)0x879844156 = 1; // write
  int var = *(volatile int*)0x879844156; // read 
  printf("var=%d\n",var);


  return 0; 
}


~/trial/pointer 307$ gcc pointer_conversion_1.c -w
~/trial/pointer 308$ ./a.out
Segmentation fault

Thanks and Regards,
Rahul Kumar

What system are you running this on?
Why do you think you should have access to this address?
What are you trying to do?

Originally Posted by algorism

What system are you running this on?
Why do you think you should have access to this address?
What are you trying to do?

It is just example code, I am trying to write at particular location/address and read back the same value from that address.

Unless you know that you can write to specific address ranges on your platform, it's a bad idea to try. Modern computers randomize addresses now. You're not supposed to care what the addresses are.

malloc() and free() still work in 90% of places.

Code:

int *var = malloc(sizeof *var);
*var = 12;
printf("*var=%d\n", *var);
free(var);

Originally Posted by rahkuma

It is just example code, I am trying to write at particular location/address and read back the same value from that address.

Even if that would work it's an insane thing to do. What if that address happened to be in the middle of your program code (and you were allowed to write to it)? It could cause your program to do anything.

You didn't bother to mention what system you are running this on, which neglect indicates Windows. But the following is true of Linux and macOS, too.

Firstly, addresses are "virtual", which means that the numeric value that your program sees are not the same as the actual underlying location in RAM. Virtual Memory

Secondly, you cannot access an arbitrary address, since only certain ranges of virtual addresses are mapped into your process. Anatomy of a Program in Memory

Thirdly, even those ranges mapped into your process will only allow certain kinds of access. Some will be read-only, for example. Some may only be executable.

Finally, as whiteflags said, for security reasons address ranges may be randomized, so you can't be sure what their actual (virtual) values will be. ASLR