lunes, diciembre 21, 2015

NcN 2015 CTF - theAnswer writeup

1. Overview

Is an elf32 static and stripped binary, but the good news is that it was compiled with gcc and it will not have shitty runtimes and libs to fingerprint, just the libc ... and libprhrhead
This binary is writed by Ricardo J Rodrigez

When it's executed, it seems that is computing the flag:

But this process never ends .... let's see what strace say:

There is a thread deadlock, maybe the start point can be looking in IDA the xrefs of 0x403a85
Maybe we can think about an encrypted flag that is not decrypting because of the lock.

This can be solved in two ways:

  • static: understanding the cryptosystem and programming our own decryptor
  • dynamic: fixing the the binary and running it (hard: antidebug, futex, rands ...)

At first sight I thought that dynamic approach were quicker, but it turned more complex than the static approach.

2. Static approach

Crawling the xrefs to the futex, it is possible to locate the main:

With libc/libpthread function fingerprinting or a bit of manual work, we have the symbols, here is the main, where 255 threads are created and joined, when the threads end, the xor key is calculated and it calls the print_flag:

The code of the thread is passed to the libc_pthread_create, IDA recognize this area as data but can be selected as code and function.

This is the thread code decompiled, where we can observe two infinite loops for ptrace detection and preload (although is static) this antidebug/antihook are easy to detect at this point.

we have to observe the important thing, is the key random?? well, with the same seed the random sequence will be the same, then the key is "hidden" in the predictability of the random.

If the threads are not executed on the creation order, the key will be wrong because is xored with the th_id which is the identify of current thread.

The print_key function, do the xor between the key and the flag_cyphertext byte by byte.

And here we have the seed and the first bytes of the cypher-text:

With radare we can convert this to a c variable quickly:

And here is the flag cyphertext:

And with some radare magics, we have the c initialized array:

radare, is full featured :)

With a bit of rand() calibration here is the solution ...

The code:

3. The Dynamic Approach

First we have to patch the anti-debugs, on beginning of the thread there is two evident anti-debugs (well anti preload hook and anti ptrace debugging) the infinite loop also makes the anti-debug more evident:

There are also a third anti-debug, a bit more silent, if detects a debugger trough the first available descriptor, and here comes the fucking part, don't crash the execution, the execution continues but the seed is modified a bit, then the decryption key will not be ok.

Ok, the seed is incremented by one, this could be a normal program feature, but this is only triggered if the fileno(open("/","r")) > 3 this is a well known anti-debug, that also can be seen from a traced execution.

Ok, just one byte patch,  seed+=1  to  seed+=0,   (add eax, 1   to add eax, 0)



To patch the two infinite loops, just nop the two bytes of each jmp $-0

Ok, but repairing this binary is harder than building a decryptor, we need to fix more things:

  •  The sleep(randInt(1,3)) of the beginning of the thread to execute the threads in the correct order
  •  Modify the pthread_cond_wait to avoid the futex()
  • We also need to calibrate de rand() to get the key (just patch the sleep and add other rand() before the pthread_create loop
Adding the extra rand() can be done with a patch because from gdb is not possible to make a call rand() in this binary.

With this modifications, the binary will print the key by itself. 

lunes, mayo 18, 2015

ASIS CTF Quals 2015 - sawthis writeup - srand remote prediction

The remote service ask for a name, if you send more than 64 bytes, a memory leak happens.
The buffer next to the name's is the first random value used to init the srand()

If we get this value, and set our local srand([leaked] ^ [luckyNumber]) we will be able to predict the following randoms and win the game, but we have to see few details more ;)

The function used to read the input until the byte \n appears, but also up to 64 bytes, if we trigger this second condition there is not 0x00 and the print shows the random buffer :)

The nickname buffer:

The seed buffer:

So here it is clear, but let's see that the random values are computed with several gpu instructions which are decompiled incorrectly:

We tried to predict the random and aply the gpu divisions without luck :(

There was a missing detail in this predcitor, but there are always other creative ways to do the things.
We use the local software as a predictor, we inject the leaked seed on the local binary of the remote server and got a perfect syncronization, predicting the remote random values:

The process is a bit ugly becouse we combined automated process of leak exctraction and socket interactive mode, with the manual gdb macro.

The macro:

Defcon 2015 coding skillz 1 writeup

Just connecting to the service, a 64bit cpu registers dump is received, and so does several binary code as you can see:

The registers represent an initial cpu state, and we have to reply with the registers result of the binary code execution. This must be automated becouse of the 10 seconds server socket timeout.

The exploit is quite simple, we have to set the cpu registers to this values, execute the code and get resulting registers.

In python we created two structures for the initial state and the ending state.

cpuRegs = {'rax':'','rbx':'','rcx':'','rdx':'','rsi':'','rdi':'','r8':'','r9':'','r10':'','r11':'','r12':'','r13':'','r14':'','r15':''}
finalRegs = {'rax':'','rbx':'','rcx':'','rdx':'','rsi':'','rdi':'','r8':'','r9':'','r10':'','r11':'','r12':'','r13':'','r14':'','r15':''}

We inject at the beginning several movs for setting the initial state:

for r in cpuRegs.keys():
    code.append('mov %s, %s' % (r, cpuRegs[r]))

The 64bit compilation of the movs and the binary code, but changing the last ret instruction by a sigtrap "int 3"
We compile with nasm in this way:

os.popen('nasm -f elf64 code.asm')
os.popen('ld -o code code.o ')

And use GDB to execute the code until the sigtrap, and then get the registers

fd = os.popen("gdb code -ex 'r' -ex 'i r' -ex 'quit'",'r')
for l in fd.readlines():
    for x in finalRegs.keys():

We just parse the registers and send the to the server in the same format, and got the key.

The code:

from libcookie import *
from asm import *
import os
import sys

host = ''
port = 9999

cpuRegs = {'rax':'','rbx':'','rcx':'','rdx':'','rsi':'','rdi':'','r8':'','r9':'','r10':'','r11':'','r12':'','r13':'','r14':'','r15':''}
finalRegs = {'rax':'','rbx':'','rcx':'','rdx':'','rsi':'','rdi':'','r8':'','r9':'','r10':'','r11':'','r12':'','r13':'','r14':'','r15':''}
fregs = 15

s = Sock(TCP)
s.timeout = 999

data = s.readUntil('bytes:')

#data =
#data = s.readAll()

sz = 0

for r in data.split('\n'):
    for rk in cpuRegs.keys():
        if r.startswith(rk):
            cpuRegs[rk] = r.split('=')[1]

    if 'bytes' in r:
        sz = int(r.split(' ')[3])

binary = data[-sz:]
code = []

print '[',binary,']'
print 'given size:',sz,'bin size:',len(binary)        
print cpuRegs

for r in cpuRegs.keys():
    code.append('mov %s, %s' % (r, cpuRegs[r]))

#print code

fd = open('code.asm','w')

print 'Compilando ...'
os.popen('nasm -f elf64 code.asm')
os.popen('ld -o code code.o ')

print 'Ejecutando ...'
fd = os.popen("gdb code -ex 'r' -ex 'i r' -ex 'quit'",'r')
for l in fd.readlines():
    for x in finalRegs.keys():
        if x in l:
            l = l.replace('\t',' ')
                i = 12
                spl = l.split(' ')
                if spl[i] == '':
                print 'reg: ',x
                finalRegs[x] = l.split(' ')[i].split('\t')[0]
                print 'err: '+l
            fregs -= 1
            if fregs == 0:
                #print 'sending regs ...'
                #print finalRegs
                buff = []
                for k in finalRegs.keys():
                    buff.append('%s=%s' % (k,finalRegs[k]))

                print '\n'.join(buff)+'\n'

                print s.readAll()
                print 'waiting flag ....'
                print s.readAll()

                print '----- yeah? -----'


lunes, marzo 30, 2015

TLS v1.2 sigalgs remote crash (CVE-2015-0291)

OpenSSL 1.0.2a fix several security issues, one of them let crash TLSv1.2 based services remotelly from internet.

Regarding to the TLSv1.2 RFC,  this version of TLS provides a "signature_algorithms" extension for the client_hello. 

Data Structures

If a bad signature is sent after the renegotiation, the structure will be corrupted, becouse structure pointer:
s->c->shared_sigalgs will be NULL, and the number of algorithms:
s->c->shared_sigalgslen will not be zeroed.
Which will be interpreted as one algorithm to process, but the pointer points to 0x00 address. 

Then tls1_process_sigalgs() will try to process one signature algorithm (becouse of shared_sigalgslen=1) then sigptr will be pointer to c->shared_sigalgs (NULL) and then will try to derreference sigptr->rhash. 

This mean a Segmentation Fault in  tls1_process_sigalgs() function, and called by tls1_set_server_sigalgs() with is called from ssl3_client_hello() as the stack trace shows.


The following code, points sigptr to null and try to read sigptr->rsign, which is assembled as movzbl eax,  byte ptr [0x0+R12] note in register window that R12 is 0x00

Debugger in the crash point.

radare2 static decompiled

The patch fix the vulnerability zeroing the sigalgslen.
Get  David A. Ramos' proof of concept exploit here