Global variables, shared libraries and -fPIC effect

Question

I made a piece of code which consists in a dynamic library (lib.c), and a main executable (main.c). In both files I define a global variable named: int global. Not very smart but it's not the question.

When I compile the dynamic library the -fPIC option seems mandatory:

gcc lib.c -fPIC -shared -o lib.so

otherwise I get:

/usr/bin/ld: /tmp/ccpUvIPj.o: relocation R_X86_64_32 against '.rodata' can not be used when making a shared object; recompile with -fPIC

When I compile the executable it is not.

gcc main.c -fPIC -ldl
gcc main.c -ldl

Both work, but have different behaviours I can not explain, could you ? :

with -fPIC, global in main.c and global in lib.c are the same variables:

global main: 23 (0x601050)
global lib: 23 (0x601050)

without -fPIC, global in lib.c is not correlated to global in main.c:

global main: 23 (0x601048)
global lib: 0 (0x7f7742e64028)

Here is the source :

lib.c

#include <stdio.h>
#include <stdlib.h>

int global;

int f_one() {

    printf("global lib: %d (%p)\n", global, &global);

    return EXIT_SUCCESS;
}

main.c

#include <stdio.h>
#include <stdlib.h>
#include <dlfcn.h>

void * handle;
int global;

int main() {

    int q = 7;

    int (* f_one_p)(int a) = NULL;

    global = 23;

    handle = dlopen("./lib.so", RTLD_NOW);

    if (handle == 0) {
        return EXIT_FAILURE;
    }

    f_one_p = dlsym(handle, "f_one");

    printf("global main: %d (%p)\n", global, &global);

    f_one_p(q);

    return EXIT_SUCCESS;

}

gcc --version: gcc (Ubuntu/Linaro 4.5.2-8ubuntu4) 4.5.2

uname -a: Linux xxx 2.6.38-11-generic #48-Ubuntu SMP Fri Jul 29 19:02:55 UTC 2011 x86_64 x86_64 x86_64 GNU/Linux

edit: code tested under SUN/sparc and x86/Linux architectures with the same kind of unexpected shared global variables (with -fPIC).

Related: http://stackoverflow.com/q/7216244/168175 – Flexo Sep 04 '11 at 09:20 — Flexo, Sep 04 '11 at 09:20

score 8 · Accepted Answer · answered Sep 04 '11 at 09:26

When you compile with -fPIC the object in question will determine the address of global symbols using the Global Offset Table. What happens though when part of the code is -fPIC and part isn't is that one of your int globals will be using this table to determine the address whilst the other part isn't.

If you had two shared object linked with -fPIC, but your main program not then you would still have two addresses for int global, one using the global offset table and one which was just local to the non-PIC code.

There's a really great discussion on PIC vs pic vs non PIC if you want to read further.

score 1 · Answer 2 · answered Sep 04 '11 at 09:27

1

By default, when building an executable references to variables are done internally, with a fixed offset and no relocation.

However, you're passing -fPIC and access to global data are converted to access via GOT and GOT relocations are added.

answered Sep 04 '11 at 09:27

Gregory Pakosz

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Global variables, shared libraries and -fPIC effect

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