I just printed function addresses and offset between two functions, but the offset is fixed whenever the program is executed (base address is only changed).

Code (test.c)

#include <stdio.h>
#include <stdint.h>

void func() {

void func2() {

int main() {
    uintptr_t main_addr = main;
    uintptr_t func_addr = func;
    uintptr_t func2_addr = func2;

    printf("main_addr: %p\n", main_addr);
    printf("func_addr: %p\n", func_addr);
    printf("func2_addr: %p\n", func2_addr);
    printf("offset (main - func): %ld\n", main_addr - func_addr);
    printf("offset (main - func2): %ld\n", main_addr - func2_addr);


gcc test.c


(1st execution)
main_addr: 0x5578d1296190
func_addr: 0x5578d129616a
func2_addr: 0x5578d129617d
offset (main - func): 38
offset (main - func2): 19

(2nd execution)
main_addr: 0x55bf299d9190
func_addr: 0x55bf299d916a
func2_addr: 0x55bf299d917d
offset (main - func): 38
offset (main - func2): 19

(3rd execution)
main_addr: 0x55ba4767d190
func_addr: 0x55ba4767d16a
func2_addr: 0x55ba4767d17d
offset (main - func): 38
offset (main - func2): 19

My kernel version is 5.4 (i.e., ubuntu 18.04)

So, my questions are:

  1. Is this expected behavior(fixed offset) in ubuntu 18.04?
  2. How can I make a situation that randomizes not only base address, but also offset.

Thank you

1 Answer 1


ASLR is about randomizing the layout of dynamically-linked libraries used by your code, not about randomizing within your program. The goal is to prevent an attacker from being able to simply hard-code the location of something like exec into a remote-execution exploit.

To illustrate, two different copies of bash from my computer have libc at different addresses according to /proc/[pid]/maps:

7fd647904000-7fd647a47000 r-xp 00022000 00:12 12970738                   /lib64/libc-2.32.so

7f2b574cd000-7f2b57610000 r-xp 00022000 00:12 12970738                   /lib64/libc-2.32.so

Instead of looking at addresses within your code, try checking the address of a library function, such as printf.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.