How to print -0x4(%rbp) in gdb?

Question

In the disassembly there's such code:

movl   $0x6,-0x4(%rbp)

I try to print the value this way but fails:

(gdb) p  0x4(%esp)
A syntax error in expression, near `%esp)'.
(gdb) p 0x4+$esp
Argument to arithmetic operation not a number or boolean.

How to print it correctly?

Paul R · Accepted Answer · 2014-10-14T21:03:20.887

21

It looks like you're working with 64 bit code, in which case you just need, e.g.:

(gdb) p $rbp-0x4
$1 = (void *) 0x7fff5fbff71c

If you want to see what's actually at this address, then you need to cast the void * address, e.g.

(gdb) p /x *(int *)($rbp-0x4)
$2 = 0x7fff

or, more succinctly, use x instead of p, e.g.

(gdb) x /w $rbp-0x4
0x7fff5fbff71c: 0x00007fff

edited Oct 14 '14 at 21:03

answered Mar 28 '11 at 08:01

Paul R

You could also use: `(gdb) x/w $rbp-0x4` – Karmastan Mar 28 '11 at 16:31
@Karmastan: cool - thanks - I'll add that as an alternative in the answer – Paul R Mar 28 '11 at 18:03
x /w * $rbp-0x4, should be x /w $rbp-0x4 – Man of One Way May 18 '12 at 09:34
1

To see what's there as a 64-bit quantity in hex: x/xg $rbp-0x4 – kaiwan Mar 18 '13 at 06:23

1 Answers1