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

解决方法：

我能够通过一次设置32位而不是一次设置所有64位来解决此问题：

(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

修改

正如@MarkPlotnick在评论中所提到的，原因和正确的分配方法是：

(gdb) whatis *0x7fffffffe338返回int，在x86_64上为32位宽。

转换为int64_t或char **将强制GDB在赋值中设置所有64位内存：

set *(int64_t *)0x7fffffffe338 = 0x7fffffffe130
或
set *(char **)0x7fffffffe338 = 0x7fffffffe130
结果

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