Here is how I captured the stripe.com CTF flag. A complete walkthrough. | Hacker News Reader

Here is how I captured the stripe.com CTF flag. A complete walkthrough.

github.com

7 pointsdividuum4 comments

I've always read about these techniques but it's way, way cooler being able to see them actually done. I don't think I would have gotten nearly this far, so such a detailed explanation of every solution taught me a lot!

Thanks a bunch

Reading this, I found it interesting how much guesswork was totally reasonable. In comparison, my solutions to the buffer overflow levels involved runnging objdump on the binary and calculating the exact indexes and addresses I wanted; the idea of "just put it in there 20 times and guess until it works" might have occurred to me, but only after I had given up understanding the specific state machine.

While this made my solution for the fun() level much cleaner (I didn't need a nop-slide, and I had no shell code: I overwrote a single address on the stack to short the program into a call to system(), return-to-libc-style), it apparently wasn't warranted to spend that time or effort: dividuum was able to metaphorically sprint through those levels while I was wasting time staring at the track. ;P

I didn't bother trying the return-to-libc approach since the bruteforce way worked for me. I guess your result is much cleaner though. Do you have any code online?

The only level that involved "code" was the final one; otherwise, it was more just running the program with the right arguments (so I'm not exactly certain what you are hoping to see). While solving the levels I tried to get each one down to an easily-paste-able bash command-line. If you log in as level04 you can just paste this command:

while ! echo 'cat /home/level05/.password' | /levels/level04 "$(printf %1036s)"$'\xd0\x23\x5b\xf7____\xee\xab\x6a\xf7' | grep -F ''; do true; done

(To be clear: I still had to brute force the ASLR; but even if you are doing the executable-stack+shell-code approach, you don't need the nop-sled as you can just calculate exactly where you intend to return to; that's what I meant by "totally reasonable" "guesswork": the nop-sled works and kept you from having to stare at the assembly.)

(Note: my original version only required a single address, but involved more messiness outside of the program; the command I pasted here is a revision I made after an e-mail exchange I had with a1k0n, who reminded me that "/bin/sh" was a string that already existed in libc, so I could run an actual shell rather than some random garbage.)

(edit: Awesome! It turns out mpetrov managed to make a 100% deterministic version, which makes the executable-stack+shell-code approach look a lot cleaner, as it drops away that irritating brute force step ;P. http://news.ycombinator.com/item?id=3630826)

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