Simple Reverse - 0x29(2023 Lab - Unpackme)

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Simple Reverse - 0x29(2023 Lab - Unpackme)

Source code

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...
LOAD:0000000000005AE8 mov     rdi, [rsp+18h+start]            ; start
LOAD:0000000000005AED push    5
LOAD:0000000000005AEF pop     rdx                             ; prot
LOAD:0000000000005AF0 push    0Ah
LOAD:0000000000005AF2 pop     rax
LOAD:0000000000005AF3 syscall                                 ; LINUX - sys_mprotect
LOAD:0000000000005AF5 jmp     r13
LOAD:0000000000005AF5
LOAD:0000000000005AF5 sub_5A7C endp
LOAD:0000000000005AF5
LOAD:0000000000005AF8 ; ---------------------------------------------------------------------------
LOAD:0000000000005AF8
LOAD:0000000000005AF8 loc_5AF8:                               ; CODE XREF: start+2↑p
LOAD:0000000000005AF8 pop     rbp
LOAD:0000000000005AF9 call    sub_5A7C
LOAD:0000000000005AF9
LOAD:0000000000005AF9 ; ---------------------------------------------------------------------------
LOAD:0000000000005AFE aProcSelfExe db '/proc/self/exe',0
LOAD:0000000000005B0D align 2
LOAD:0000000000005B0E dw 1
LOAD:0000000000005B10 dq 81B00000C1100h, 0FFFFFF0000000200h, 7549F983004AE8E5h, 0FD374C8D48575344h, 0CE39482FEB5B565Eh, 0FFFFFBFF5E563273h
LOAD:0000000000005B10 dq 778F3C0A72803CACh, 2C06740FFE7E8006h, 56161BE477013CE8h, 0FFBFFFFF75D028ADh, 0D801F829C80F5FDFh, 0C35BDFEBAC0312ABh
LOAD:0000000000005B10 dq 8948505741564158h, 0DBFFEDFEEC8148E6h, 590A6A5F54591000h, 5003E8348A548F3h, 0B6AB48FE8949F875h, 0F60C0AFC0CCBB374h
LOAD:0000000000005B10 dq 4DF5FF6EDFFE02FFh, 5E57370FFFBAFC29h, 50F58596AED7B8Ch, 0DFFF6FDB0579C085h, 8D49FD91580F6A0Eh, 0E741AAA00B0FF7Dh
...
  • Real File main Function 
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    int __cdecl main(int argc, const char **argv, const char **envp)
{
  int result; // eax
  unsigned __int64 i; // [rsp+8h] [rbp-58h]
  char user_input[32]; // [rsp+10h] [rbp-50h] BYREF
  char v6[40]; // [rsp+30h] [rbp-30h]
  unsigned __int64 v7; // [rsp+58h] [rbp-8h]

  v7 = __readfsqword(0x28u);
  printf("Enter input: ");
  scanf("%s", user_input);
  if ( sub_10C0(user_input, qword_4018, 10LL) )
  {
    printf_0("Incorrect!");
    result = 1;
  }
  else
  {
    for ( i = 0LL; i <= 0x26; ++i )
      v6[i] = user_input[i % 0xA] ^ *(qword_4010 + i);
    printf("%s");
    result = 0;
  }
  if ( v7 != __readfsqword(0x28u) )
    return sub_10A0();
  return result;
}

Recon

這一題一開始就知道是UPX加殼，但是直接試了upx幫忙decompress，卻遇到error，代表可能有一些問題(在Unix環境底下?)，所以我嘗試使用手動脫殼，去分析其中的內容

  1. 首先可以先靜態看一下脫完殼之前是在哪邊跳轉，經過實測和判斷，應該是: 
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     LOAD:0000000000005AF5 jmp     r13

    如何在動態取得這一行的位置呢?手動算出rebase address

    1. 首先先用靜態分析看starti的時候的offset
    2. 開始動態執行程式
    3. 把目前指到的address拿去和靜態分析拿到的offset相減
    4. (optional)可以用vmmap確認一下
    5. 再把我們想要得知的那一行的offset加回來

    圖片.png 一開始的offset是0x5888

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     gef➤  starti
 gef➤  x/x 0x7ffff7ffd888-0x5888
 0x7ffff7ff8000: 0x7f
 gef➤  vmmap
 [ Legend:  Code | Heap | Stack ]
 Start              End                Offset             Perm Path
 0x00007ffff7ff2000 0x00007ffff7ff6000 0x0000000000000000 r-- [vvar]
 0x00007ffff7ff6000 0x00007ffff7ff8000 0x0000000000000000 r-x [vdso]
 0x00007ffff7ff8000 0x00007ffff7ff9000 0x0000000000000000 rw- /mnt/d/NTU/Second Year/Computer Security/Reverse/Lab3/Unpackme/unpackme
 0x00007ffff7ff9000 0x00007ffff7ffd000 0x0000000000000000 rw-
 0x00007ffff7ffd000 0x00007ffff7fff000 0x0000000000000000 r-x /mnt/d/NTU/Second Year/Computer Security/Reverse/Lab3/Unpackme/unpackme
 0x00007ffffffdd000 0x00007ffffffff000 0x0000000000000000 rw- [stack]
 gef➤  x/10i 0x7ffff7ff8000+0x5AF5
     0x7ffff7ffdaf5:      jmp    r13
     0x7ffff7ffdaf8:      pop    rbp
     0x7ffff7ffdaf9:      call   0x7ffff7ffda7c
     0x7ffff7ffdafe:      (bad)
     0x7ffff7ffdaff:      jo     0x7ffff7ffdb73
     0x7ffff7ffdb01:      outs   dx,DWORD PTR ds:[rsi]
     0x7ffff7ffdb02:      movsxd ebp,DWORD PTR [rdi]
     0x7ffff7ffdb04:      jae    0x7ffff7ffdb6b
     0x7ffff7ffdb06:      ins    BYTE PTR es:[rdi],dx
     0x7ffff7ffdb07:      data16 (bad)
  2. 利用動態看r13的address會跳去哪邊 → 0x00007ffff7ff1000

  3. 接下來我找不太到分析的地方，所以就直接c(continue)到user input的地方停下來，再看vmmap

    vmmap

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     [ Legend:  Code | Heap | Stack ]
 Start              End                Offset             Perm Path
 0x00007ffff7d84000 0x00007ffff7d87000 0x0000000000000000 rw-
 0x00007ffff7d87000 0x00007ffff7daf000 0x0000000000000000 r-- /usr/lib/x86_64-linux-gnu/libc.so.6
 0x00007ffff7daf000 0x00007ffff7f44000 0x0000000000028000 r-x /usr/lib/x86_64-linux-gnu/libc.so.6
 0x00007ffff7f44000 0x00007ffff7f9c000 0x00000000001bd000 r-- /usr/lib/x86_64-linux-gnu/libc.so.6
 0x00007ffff7f9c000 0x00007ffff7fa0000 0x0000000000214000 r-- /usr/lib/x86_64-linux-gnu/libc.so.6
 0x00007ffff7fa0000 0x00007ffff7fa2000 0x0000000000218000 rw- /usr/lib/x86_64-linux-gnu/libc.so.6
 0x00007ffff7fa2000 0x00007ffff7faf000 0x0000000000000000 rw-
 0x00007ffff7fb3000 0x00007ffff7fb5000 0x0000000000000000 rw-
 0x00007ffff7fb5000 0x00007ffff7fb7000 0x0000000000000000 r-- /usr/lib/x86_64-linux-gnu/ld-linux-x86-64.so.2
 0x00007ffff7fb7000 0x00007ffff7fe1000 0x0000000000002000 r-x /usr/lib/x86_64-linux-gnu/ld-linux-x86-64.so.2
 0x00007ffff7fe1000 0x00007ffff7fec000 0x000000000002c000 r-- /usr/lib/x86_64-linux-gnu/ld-linux-x86-64.so.2
 0x00007ffff7fec000 0x00007ffff7fed000 0x0000000000000000 ---
 0x00007ffff7fed000 0x00007ffff7fef000 0x0000000000037000 r-- /usr/lib/x86_64-linux-gnu/ld-linux-x86-64.so.2
 0x00007ffff7fef000 0x00007ffff7ff1000 0x0000000000039000 rw- /usr/lib/x86_64-linux-gnu/ld-linux-x86-64.so.2
 0x00007ffff7ff2000 0x00007ffff7ff6000 0x0000000000000000 r-- [vvar]
 0x00007ffff7ff6000 0x00007ffff7ff8000 0x0000000000000000 r-x [vdso]
 0x00007ffff7ff8000 0x00007ffff7ff9000 0x0000000000000000 r--
 0x00007ffff7ff9000 0x00007ffff7ffa000 0x0000000000000000 r-x
 0x00007ffff7ffa000 0x00007ffff7ffc000 0x0000000000000000 r--
 0x00007ffff7ffc000 0x00007ffff7ffd000 0x0000000000000000 rw-
 0x00007ffff7ffe000 0x00007ffff7fff000 0x0000000000000000 r-- /mnt/d/NTU/Second Year/Computer Security/Reverse/Lab3/Unpackme/unpackme
 0x00007ffff7fff000 0x00007ffff8020000 0x0000000000000000 rw- [heap]
 0x00007ffffffdd000 0x00007ffffffff000 0x0000000000000000 rw- [stack]

    可以看到0x00007ffff7ff8000開始會有ELF的字樣，代表應該是他脫殼完的結果，我的作法是直接把0x00007ffff7ff8000~0x00007ffff7ffd000全部dump下來進行分析

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     gef➤  x/s 0x00007ffff7ff8000
 0x7ffff7ff8000: "\177ELF\002\001\001"
 gef➤  dump memory real_file 0x00007ffff7ff8000 0x00007ffff7ffd000

  4. 開始分析real_file，先用靜態看一下(如source code所示) 圖片.png

  5. 找到我們要停的地方的offset → 0x1213 
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     gef➤  x/10i 0x00007ffff7ff8000+0x1213
     0x7ffff7ff9213:      mov    rcx,QWORD PTR [rip+0x2dfe]        # 0x7ffff7ffc018
     0x7ffff7ff921a:      lea    rax,[rbp-0x50]
     0x7ffff7ff921e:      mov    edx,0xa
     0x7ffff7ff9223:      mov    rsi,rcx
     0x7ffff7ff9226:      mov    rdi,rax
 => 0x7ffff7ff9229:      call   0x7ffff7ff90c0
     0x7ffff7ff922e:      test   eax,eax
     0x7ffff7ff9230:      je     0x7ffff7ff924b
     0x7ffff7ff9232:      lea    rax,[rip+0xe13]        # 0x7ffff7ffa04c
     0x7ffff7ff9239:      mov    rdi,rax

    可以看到解析出來的assembly和IDA的差不多，代表我們找對地方

  6. 設定breakpoint後continue就可以在stack中看到key 
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     gef➤  b *(0x00007ffff7ff9000+0x229)
 Breakpoint 1 at 0x7ffff7ff9229
 gef➤  c
 Continuing.
 adjfl

 Breakpoint 1, 0x00007ffff7ff9229 in ?? ()

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     ────────────────────────────────────────────────────────────────────────────────────────── arguments (guessed) ────
 0x7ffff7ff90c0 (
     $rdi = 0x00007fffffffd6c0 → 0x0000006c666a6461 ("adjfl"?),
     $rsi = 0x00007ffff7ffa030 → "just_a_key",
     $rdx = 0x000000000000000a,
     $rcx = 0x00007ffff7ffa030 → "just_a_key"
 )

Exploit

key: just_a_key

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$ ./unpackme
Enter input: just_a_key
FLAG{just_4_simple_unpackme_challenge!}

Flag: FLAG{just_4_simple_unpackme_challenge!}