Re: Viewing the address of an array in gdb

On 07/06/2018 01:00 PM, Pedro Alves wrote:
> On 07/06/2018 12:51 PM, Pedro Alves wrote:
>> On 07/06/2018 04:33 AM, Mahmood Naderan via gdb wrote:
>>> Hi,
>>> I have define "char buffer[100]" in a C code. Trying to view the location of buffer in GDB, I see this
> 
> Ah, I missed the "you have defined a buffer yourself" part.
> 
> So what seems to be happening is that there's another symbol also 
> called "buffer" in glibc:
> 
>  (gdb) info symbol buffer
>  buffer in section .bss of /lib64/libc.so.6
> 
> and gdb is picking that symbol instead of yours.
> 
> I can reproduce this if I compile a small program without debug info,
> but with debug info (-g), it works as you'd expect.
> 
> I don't think we have syntax to disambiguate this, like
> 'libc.so.6'::buffer' vs 'program'::buffer, or program#buffer some
> such, unfortunately.
If you can't change the program, you can work around this by
unloading the shared library symbols, so that gdb sees the
program's symbol again.  E.g.:

(gdb) start
Temporary breakpoint 1 at 0x40048b
Starting program: /home/pedro/tmp/buffer 

Temporary breakpoint 1, 0x000000000040048b in main ()
(gdb) info symbol &buffer
buffer in section .bss of /lib64/libc.so.6
(gdb) nosharedlibrary 
(gdb) info symbol &buffer
buffer in section .bss
(gdb) x /30x  &buffer
0x601040 <buffer>:      0x00000000      0x00000000      0x00000000      0x00000000
0x601050 <buffer+16>:   0x00000000      0x00000000      0x00000000      0x00000000
0x601060 <buffer+32>:   0x00000000      0x00000000      0x00000000      0x00000000
0x601070 <buffer+48>:   0x00000000      0x00000000      0x00000000      0x00000000
0x601080 <buffer+64>:   0x00000000      0x00000000      0x00000000      0x00000000
0x601090 <buffer+80>:   0x00000000      0x00000000      0x00000000      0x00000000
0x6010a0 <buffer+96>:   0x00000000      0x00000000      0x00000000      0x00000000
0x6010b0:       0x00000000      0x00000000
(gdb) 

Thanks,
Pedro Alves