Question

在拆解中,有这样的法典:

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

我试图以这种方式印刷价值,但未能做到:

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

如何正确印刷?

Answer 1

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

如果你想看到这一地址实际上是什么at,那么,你需要加入,避免* 地址,例如:

(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