[SOLVED] why does memory address change in Ubuntu and not in Redhat

Issue

I have this program:

double t;
main() {
}

On Ubuntu, I run:

% gdb a.out
(gdb) p &t
$1 = (double *) 0x4010 <t>
(gdb) run
Starting program: /home/phan/a.out 
[Inferior 1 (process 95930) exited normally]
(gdb) p &t
$2 = (double *) 0x555555558010 <t>

Why did the address change from 0x4010 to 0x555555558010. Is there someway to prevent this? On Redhat, it doesn't do that:

% gdb a.out
(gdb) p &t
$1 = (double *) 0x601038 <t>
(gdb) r
Starting program: /home/phan/a.out 
[Inferior 1 (process 23337) exited normally]
(gdb) p &t
$2 = (double *) 0x601038 <t>

BTW, this only occurs in Ubuntu 18.04. In Ubuntu 16.04, it works exactly as Redhat, ie the address is the same before and after.

Solution

You are presumably seeing pre and post-relocation addresses for the .bss segment.

You can avoid this by disabling position independent executables, thus making gcc choose the final address of the .bss register up front:

gcc -no-pie foo.c

-static would have the effect.

I don't know why there'd be a difference between Ubuntu and Redhat though.

Answered By - that other guy

Answer Checked By - David Goodson (WPSolving Volunteer)