how objdump handles global variables

Question

I have made the following dummy code for testing

/tmp/test.c contains the following:

#include "test.h"
#include <stdio.h>
#include <stdlib.h>
struct s* p;
unsigned char *c;

void main(int argc, char ** argv) {
   memset(c, 0, 10);
   p->a = 10;
   p->b = 20;
}

/tmp/test.h contains the following:

struct s {
    int a;
    int b;
};

I compile and run objdump as follows:

cd /tmp gcc -c test.c -o test.o objdump -gdsMIntel test.o

I get the following output:

test.o:     file format elf32-i386

Contents of section .text:
 0000 5589e5a1 00000000 c7000000 0000c740  U..............@
 0010 04000000 0066c740 080000a1 00000000  .....f.@........
 0020 c7000a00 0000a100 000000c7 40041400  ............@...
 0030 00005dc3                             ..].            
Contents of section .comment:
 0000 00474343 3a202855 62756e74 752f4c69  .GCC: (Ubuntu/Li
 0010 6e61726f 20342e36 2e332d31 7562756e  naro 4.6.3-1ubun
 0020 74753529 20342e36 2e3300             tu5) 4.6.3.     
Contents of section .eh_frame:
 0000 14000000 00000000 017a5200 017c0801  .........zR..|..
 0010 1b0c0404 88010000 1c000000 1c000000  ................
 0020 00000000 34000000 00410e08 8502420d  ....4....A....B.
 0030 05700c04 04c50000                    .p......        

Disassembly of section .text:

00000000 <main>:
   0:   55                      push   ebp
   1:   89 e5                   mov    ebp,esp
   3:   a1 00 00 00 00          mov    eax,ds:0x0 ;;;; should be the address of unsigned char *c
   8:   c7 00 00 00 00 00       mov    DWORD PTR [eax],0x0 ;;;; setting 10 bytes to 0
   e:   c7 40 04 00 00 00 00    mov    DWORD PTR [eax+0x4],0x0
  15:   66 c7 40 08 00 00       mov    WORD PTR [eax+0x8],0x0
  1b:   a1 00 00 00 00          mov    eax,ds:0x0
  20:   c7 00 0a 00 00 00       mov    DWORD PTR [eax],0xa ;;;; p->a = 10;
  26:   a1 00 00 00 00          mov    eax,ds:0x0
  2b:   c7 40 04 14 00 00 00    mov    DWORD PTR [eax+0x4],0x14 ;;;; p->b = 20;
  32:   5d                      pop    ebp
  33:   c3                      ret

In the above disassembly, I find that:

In the case of c, the following is done:

mov eax, ds:0x0
mov DWORD PTR [eax], 0

In the case of p->a the following is done:

mov    eax, ds:0x0
mov    DWORD PTR [eax],0x0

In that case, are both c and p->a located in the same address (ds:0x0)?

Well, as per my understanding after reading [this link](http://www.ivor.it/cle266/guide.html), the addresses are actually relocation addresses which are printed as relocated addresses when objdump is used with -r option (-R is for .so's, -r is for .o's) — Deaf Ear, Aug 02 '12 at 06:19
Possible duplicate of [objdump and resolving linkage of local function calls?](https://stackoverflow.com/questions/8992938/objdump-and-resolving-linkage-of-local-function-calls) — Peter Cordes, Feb 11 '19 at 05:07
Use `objdump -drwC -Mintel` (`-r` being the key) to print relocations. — Peter Cordes, Feb 11 '19 at 05:08

score 1 · Answer 1 · answered Nov 21 '12 at 16:54

The .o file is not yet executable, there'll be relocation entries pointing to those 0's which the linker will fixup later. Since your source file seems mostly self-contained, can you drop the -c flag to produce an (fully linked) executable rather than a relocatable object file?

how objdump handles global variables

1 Answers1