Continuing the series of posts about the FlareOn reversing challenges 2015
]. If you interested, you can find my previous post with solutions for the first three challenges here
I had some "extra work" due to infra-structure issues during this challenge and ended up learning new things. So I decided to dedicate a whole post for the challenge #4 not only explaining the solution but also explaining these issues.
As the previous challenges the objective in this task is to recover the flag (an e-mail
address) from a PE-32
binary file. The main insight for this challenge is that the binary was packed using UPX
]. There are many ways to raise evidences that a binary was packed using UPX
, the main is to examine sections. We could obtain this information using rabin2
by the following command. Figure 01 shows the sections labels for the analyzed binary.
|Figure 01: Sections information for analyzed binary.|
As we can notice there are 2
sections (marked with the red rectangles) called "UPX0
" and "UPX1
". These sections are used by UPX
to store compressed data of the original binary.UPX
standard package offers an utility to unpack the packed binary. Basically this tool reads the UPX
headers (out of the scope for this post) and extracts the original binary to an output file. The main advantage of using UPX
is to reduce the size of an executable by compacting it. This compressed version of the binary is uncompressed in memory and executed during run-time. This means that the packed executable has the decompression algorithm and a mechanism to change the execution flow to the uncompressed version of the binary during execution time.
.::[ UNPACKING UPX
At this point we already know that we have an UPX packed binary. The binary executes perfectly packed at our test environment (Microsoft Windows 7 - 64 bits
). The issues happens when we tried to unpacking and executing the resulting unpacked binary using this Virtual Machine. Surprisedly the resulting unpacked binary does not work anymore. Figure 02 shows the command line used to unpack the challenge and the checksum for the unpacked version of the binary.
|Figure 02: Command line for unpacking binary and MD5|
Figure 03 shows the error presented when we tried to execute the unpacked version of the binary.
|Figure 03: Execution error of unpacked version of the binary|
This was very curious, we analyzed the unpacking code and the uncompressing routine was fine. Then we decided to execute the unpacked binary in different platforms, such as:
- Microsoft Windows 7 - 32 bits
- Microsoft Windows XP - 32 bits
The same error occurs when we tried to execute on Windows 7 - 32 bits
. The unpacked binary was executed without any error on Windows XP - 32 bits
. Figure 04 shows the execution of the unpacked binary under a Windows XP - 32 bits
|Figure 04: Executing the unpacked version of the binary in a Windows XP - 32 bits|
This situation raises the Issue investigated in this post (a little bit off-topic but interesting though):
"why the unpacked version of the binary executes on Windows XP but not on Windows 7 even though the packed version executes without any error on both platforms?"
.::[ ISSUE ANALYSIS
After loading the unpacked version of the binary on IDA we could notice that the executable was compiled by using some version of Visual C with the flag "GS
" active [4
]. This flag tells the compiler to inject code to check the software against buffer overflows. The code implements a set of cookies
based routines to check if the boundaries of variables are not violated. Figure 05 shows the code inserted by the compiler used to generate this security cookie
Why we are talking about this? Because the debugger reports a memory access violation when the program tries read the security cookie initialized and stored inside the data segment by the function "__init_security_cookie()"
. Figure 06 shows the exception triggered by the debugger and the defective code.
|Figure 06: Exception triggered when executing the unpacked binary.|
An interesting behavior is that the code is modified when executed from:
This runtime change is causing the error because the register DS
(Data Segment register
) is not initialized with the base address for the Data Segment
. Well, the Data Segment Register
should be filled dynamically according the Relocation Table
] and is used for ASLR
]. The problem is that the binary does not have Relocation Table
(probably striped by UPX
) and was compiled with the ASLR
flag. This information can be confirmed by looking at the PE structure by using CFF Explorer
] and Process Explorer
]. Figure 07 shows the information about ASLR
flag at Process Explorer
. As we can see the process has the ASLR flag active.
|Figure 07: Process characteristics interface showing that ASLR flag is active |
We can double check and edit characteristics of the binary by using CFF Explorer
. Figure 08 shows this same information ASLR
flag on CFF Explorer
but this time this tool was used to unset this flag.
|Figure 08: unpacked binary characteristics on CFF Explorer |
Once the flag was disabled the binary does not uses ASLR
and does not requires the Relocation Table
anymore. We can observe now on Figure 09 that the unpacked binary is executing normally on Windows 7
|Figure 09: Unpacked binary is executing fine after ASLR flag was disable|
The unpacked binary was executing normally on Windows XP
because this version of Windows Operating System
does not implement ASLR
was implemented for the first time on Windows Vista
.::[ CHALLENGE 04 - HIGH LEVEL BEHAVIOR
No more obscure errors, now we can finally focus on the challenge. The first thing to notice in this challenge was the presence of UPX
. Another interesting characteristic is the difference of output between the unpacked and the packed version. The only reason for such thing is that the unpacking algorithm is modifying the unpacked binary during execution time. After analyzing the end of the algorithm used for unpacking the original code, just before redirecting the execution flow for the Original Entry Point
, some routines was injected to modify the unpacked binary. There are 2
- replaces a "5" by "4" in a specific position of the memory;
- switch the cases of an array of characters ("A" goes to "a", "B" to "b" and so on) .
Figure 10 shows the changes made by the modified version of the UPX unpacking algorithm over the packed binary. It is possible also to see the jump to the Original Entry Point
at address "0x408621
" this instruction changes the execution flow to the beginning of the UPX0
|Figure 10: changes made by the modified version of the UPX unpacking algorithm over the packed binary|
In order to use the unpacked version of the binary for analyzing the challenge it is necessary to manually patch the unpacked binary and add the modification code injected inside the unpacking algorithm.
The first patch can be achieved by changing the byte at the offset "0x3c4c
" from "0x35
" (string "5
") to "0x34
" (string "4
"). The second patch can be achieved by changing the content starting at the offset "0x3bb8
" until offset "0x3beb
". It is necessary to switching the case of all alphabetic characters ("A
" goes to "a
" to "b
" etc). Figure 11 shows the changes for the second patch.
|Figure 11: second patch which swap the case of the characters for a specific region in memory.|
By now we have restored the original binary and we are ready to start to analyze what is necessary to solve the challenge.
The first thing to notice is the verification code to check if the binary was unpacked properly. This code checks if the static string located inside "0x0040525c
" (labeled as "Str
" by IDA) is not equal to "5
". Remembering that the customized unpacking algorithm replaces this string with "4
" during execution time. The routine calls the function "atoi()
" to transform this string in a integer and makes the comparison. Figure 12 shows the unpacking checking routine.
|Figure 12: unpacking checking routine|
The second thing to notice is that the binary expects a parameter. Figure 13 shows the routine which checks the number of parameters passed to the binary.
|Figure 13: Routine which checks the number of parameters passed to the binary|
The third relevant code is the call to the the function "localtime64()"
which returns a structure containing the current time at the system. Figure 14 shows the code which calls this routine. Another interesting thing to notice at this routine is a code to calculate a MD5
hash of the inputed parameter (first red rectangle). The function "sub_4012E0
" calculates this MD5
|Figure 14: localtime call routine.|
At this point we know that the binary receives an integer
parameter with maximum 8 bits (because the mov instruction receiving "al
"). Well we could just ignore the rest of the code and brute force the binary with values between 0
. This can give us the flag and solve the challenge. But our target is understanding the behave of the code then we should move on.
The next characteristic is that the binary collects the local hour at the system and user this integer to as an index for an array containing 24
strings encoded with Base64
. Inside each component of this array has an encoded MD5 hash
. This hash is compared with the MD5 hash
calculated by using the input argument; if it does not match
the binary exits. Figure 15 shows the code responsible for collecting the hour in the host system.
|Figure 15: collecting the hour in the host system|
Figure 16 shows the function responsible for using the current hour for indexing the array with the hashes in Base64
|Figure 16: accessing the "hashed" hours vector |
The function in "sub_401000
" is responsible for decoding the content and compare the decoded hash with the one calculated by using the current hour. This is the necessary information to solve the challenge, we know that the binary expects as parameter the current hour of the host system. Figure 17 shows the resulting output by inserting the current hour as first parameter to the binary.
|Figure 17: Binary executed with the current hour of the system as a parameter.|
The challenge is solved but we still want to understanding the rest of the analyzed binary. Well by observing the graph generated by IDA we can observe that the _main function flow structure can give us some hints. Figure 18 shows the flow graph of the main function calculated by IDA. We can observe that the main function has basically 2
sequences of code with 24
functional blocks each.
|Figure 18: Flow graph of the _main function generated by IDA.|
The first set of blocks is responsible to building the vector containing the encoded hashes for numbers between 1
. This is used to verify if the user is inserting the expected time as a parameter to the binary. The second set of blocks builds another vector containing an encrypted version of the flag for each hour. The hash representation of the hour is used as a XOR
key for decrypting the flag. Figure 19 shows the final decryption routine which is used to recover the flag.
|Figure 19: XOR decryption routine|
That's all folks!
PS: during my journey to understanding how relocation tables works, I found a very interesting article about code injection
 Windows Internals (Chapter #4)
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