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Ringzer0team CTF Writeup – ELF Crackme 1 – Time to Learning x86 ASM & gdb

Register yourself if you have not yet.

Access the challenge at http://ringzer0team.com/challenges/11

Download the subject,  the checksum:

  • f6816b590d2021a16ba8005aa235e6a3 (md5)
  • b8c18db3e4678e09683e3b20e9004d1183c2420b (sha1)

The challenge clearly instruct as to utilize GDB. In this case, I have customize my GDB using init script which can be downloaded from my github.

After downloading the binary, ran ‘file’ then ‘readelf’ to get some initial information about the file.

# file 88eb31060c4abd0931878bf7d2dd8c1a
88eb31060c4abd0931878bf7d2dd8c1a: ELF 32-bit LSB  executable, Intel 80386, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.24, BuildID[sha1]=a5f44b829c4727ed369f823f19d575087673f34e, not stripped

# readelf -h 88eb31060c4abd0931878bf7d2dd8c1a
ELF Header:
  Magic:   7f 45 4c 46 01 01 01 00 00 00 00 00 00 00 00 00 
  Class:                             ELF32
  Data:                              2's complement, little endian
  Version:                           1 (current)
  OS/ABI:                            UNIX - System V
  ABI Version:                       0
  Type:                              EXEC (Executable file)
  Machine:                           Intel 80386
  Version:                           0x1
  Entry point address:               0x8048380
  Start of program headers:          52 (bytes into file)
  Start of section headers:          4508 (bytes into file)
  Flags:                             0x0
  Size of this header:               52 (bytes)
  Size of program headers:           32 (bytes)
  Number of program headers:         9
  Size of section headers:           40 (bytes)
  Number of section headers:         30
  Section header string table index: 27

We know the entrypoint which is 0x8048380 and certain that the file is ELF32.

Load the binary to GDB, we use Intel syntax instead of AT&T syntax and break to entrypoint. We then run the binary so we can reach our breakpoint.

# gdb 88eb31060c4abd0931878bf7d2dd8c1a
gdb$ set disassembly-flavor intel
gdb$ break *0x8048380
gdb$ run

If you are using my .gdbinit script you can see the all the registers. If not, see the disassembly of $eip and let’s analyze the code.

gdb$ disassemble $eip

See the code and learn that there are interesting parts.

...
   0x080484ae <+66>:	mov    DWORD PTR [eax],0x47414c46
   0x080484b4 <+72>:	mov    DWORD PTR [eax+0x4],0x3930342d
   0x080484bb <+79>:	mov    WORD PTR [eax+0x8],0x32
...
   0x080484e9 <+125>:	mov    DWORD PTR [eax],0x75393438
   0x080484ef <+131>:	mov    DWORD PTR [eax+0x4],0x6a326f69
   0x080484f6 <+138>:	mov    WORD PTR [eax+0x8],0x66
...
   0x08048530 <+196>:	mov    DWORD PTR [eax],0x6a736c6b
   0x08048536 <+202>:	mov    DWORD PTR [eax+0x4],0x6c6b34

All of them are pushing the code into some region of memory pointed by eax. It’s ASCII, if you know. Let’s search it in our ASCII table.

0x47414c46   ==> GALF
0x3930342d   ==> 904-
0x32         ==> 2
...
0x75393438   ==> u948
0x6a326f69   ==> j2oi
0x66         ==> f
...
0x6a736c6b   ==> jslk
0x6c6b34     ==> lk4

Well, it doesn’t make sense, unless you remember they are pushed in reverse order for each word. So, the flag would be

FLAG-4092849uio2jfklsj4kl

Submit the flag!

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