How to set an arbitrary 64 bits of memory in GDB

Question

I'm learning to use GDB on my own (for the purpose of understanding an assignment that requires binary analysis) and need a little bit of help. I've looked through the manual but can't find an answer to this specific issue.

I know where a 64-bit pointer resides in memory, and I want to change the address that it points to. When I try to set the value of the memory address, it only seems to modify the last 32 bits instead of the entire 64 bits.

(gdb) x/xg $rbp-8
0x7fffffffe338: 0x0000000000400a2d
(gdb) set *0x7fffffffe338 = 0x7fffffffe130
(gdb) x/xg $rbp-8
0x7fffffffe338: 0x00000000ffffe130

What's going on here?

Thanks in advance!

Answer 1

Workaround:

I was able to work around this issue by setting 32 bits at a time rather than all 64 bits at once:

(gdb) x/xg $rbp-8
0x7fffffffe548: 0x0000000000400a2d

#little-endian
(gdb) set *0x7fffffffe548 = 0xffffe130
(gdb) set *0x7fffffffe54c = 0x00007fff

(gdb) x/2xw $rbp-8
0x7fffffffe548: 0xffffe130      0x00007fff
(gdb) x/xg $rbp-8
0x7fffffffe548: 0x00007fffffffe130

EDIT:

As mentioned by @MarkPlotnick in the comments, the reason and correct method of assignment for this is:

(gdb) whatis *0x7fffffffe338 returns int, which is 32-bits wide on x86_64.

Casting to int64_t or char** will force GDB to set all 64-bits of memory in the assignment:

set *(int64_t *)0x7fffffffe338 = 0x7fffffffe130
or
set *(char **)0x7fffffffe338 = 0x7fffffffe130
results in

(gdb) x/xg $rbp-8
0x7fffffffe548: 0x00007fffffffe130