CODEWITHSUNDEEP
cstackbufferoverflow
Sean K
Sean K

Reputation: 147

Buffer overflow issue: array is shorter than it should be?

I'm working on an exercise in my systems programming class right now that deals with buffer overflows. Since I can't get the problem statement formatted correctly, I'll just paraphrase it. We have a buffer that is size 512 chars. A function named getbufn is called that allocates this buffer and then calls the Gets() function to receive input. The input comes in the form of hex values in a text document separated by spaces that is ran through another provided program to produce the input. Once the input is inputted into the buffer by gets, getbufn sets eax = 1 and returns to a test function that checks to see if the stack has been corrupted.

Now I've already gotten this to work before. In the last problem the buffer was size 32 and the stack was set. Now the stack can move. The code that calls getbufn first allocates a random amount of storage on the stack, such that if you sample the value of %ebp during two successive executions of getbufn, you would find they differ by as much as ±240. So we need to use a NOP sled to make our code work.

First off, here is the getbufn function:

80491e8:       55                      push   %ebp
80491e9:       89 e5                   mov    %esp,%ebp
80491eb:       81 ec 18 02 00 00       sub    $0x218,%esp
80491f1:       8d 85 f8 fd ff ff       lea    -0x208(%ebp),%eax
80491f7:       89 04 24                mov    %eax,(%esp)
80491fa:       e8 db fa ff ff          call   8048cda <Gets>
80491ff:       b8 01 00 00 00          mov    $0x1,%eax
8049204:       c9                      leave  
8049205:       c3                      ret    
8049206:       90                      nop
8049207:       90                      nop

I've found a few things out. When ret is called at address 0x8049205, the esp = 0x556832F4. So I know the return address is there. At the mov instruction after lea, ebp always equals 0x556832F0 (so I don't know what my prof means by it moves by 240), and eax always equals 0x556830e8. This, to me, means that my array starts at 0x556830e8 and ends at 0x556832E8. Then to get the return address, I should have 12 more bytes that I need to write to get to 0x556832F4. Then the final 4 bytes should be an address in the middle of the NOP sled.

So right now what I have is a ~500 NOP OPS and then my code at the end of the buffer - which makes the total size 512 to fill up the array. Then I have 12 bytes of hex that match up to what was originally on the stack at these addresses before the buffer overflow. Then I have 4 more bytes that points to an address in the middle of the NOP sled when the return instruction executed at the end of getbufn. This isn't working for some reason. If I put 5 bytes of garbage between my array and the return address, I just get that the stack was corrupted. If I put 6 bytes of garbage between my array and the return value, I get a segmentation fault. In my calculations, I thought I needed 12 bytes to get from my array to the return address, so what am I doing wrong here?

Upvotes: 2

Views: 292

Answers (1)

Scotty Bauer
Scotty Bauer

Reputation: 1277

This sounds a lot like the CS:APP buffer overflow lab from Bryant and O'Hallaron, it also sounds like the last level.

Let me give you some pointers:

  1. The stack is still executable.
  2. You know the buffer is ~512bytes, (may be more for alignment, when I did this assignment it was 540 bytes)
  3. You know $ebp won't differ more than +/- 240 bytes.
  4. You know a function will be called to validate the stack

So knowing all this you can do the following:

  1. Put shell code in that 512byte buffer.
  2. Some NOPS which will allow you to jump to some random point into the buffer( this is important due to the +/- 240 ASLR)
  3. In that buffer you can have some shell code that reads $esp and adds 512 bytes (or w/e the real size of th e buffer is) to $esp and that will get you the value of the saved $ebp you overwrote. You can then do some calculations and replace the garbage ebp with the correct one you just calculated.
  4. return to where ever you need to be.

To explain some things further. Why do we need NOPs (\x90) in the buffer? Since the base address gets randomized by +/- 240bytes and the buffer size is 512 bytes you can do a calculation and deterministically always return to somewhere in the buffer. When you return to somewhere in the buffer, it will hit the NOPS (\x90) and slide into your shell code.

If you have any further questions, feel free to ask.

Upvotes: 0

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