5 ms·

This paper was written 7 years before SELinux was merged.

• sown 17y ago
  Yes. And the examples won't always work on modern linux distros depending on configuration
  • petsos 17y ago
    Well, I would hope so that exploiting techniques written 13 years ago won't always work on modern distros.
    • jacquesm 17y ago
      Your hope is unfortunately misplaced, the 'always' is the operative word. Probably we will never see the end of this kind of attack until we switch to some radically different architecture.
      • jrockway 17y ago
        "Managed code."
        • swolchok 17y ago
          Still subject to attacks on the management system, which is written with existing architecture. True Harvard architecture instead of Von Neumann may help, as it discriminates between code and data.
        • jacquesm 17y ago
          That's limited to the microsoft platform for now isn't it ? Or does the term also apply to java ? One of the things that I've been wondering about, that maybe you can answer is this: If languages such as javascript, lisp, python and so on (as opposed to say C) treat functions as first class citizens, doesn't that open up a completely new can of worms with respect to security ?
          • tptacek 17y ago
            A strange question. Almost every language we use today is "managed", by the Microsoft definition; "Managed C++" is a Microsoft term for bare-metal C++ with extra runtime protections. Of course Java is "managed".
            • jrockway 17y ago
              Yes, "managed C++" is very poorly managed. You can still do things like: char foo[42]; foo[-4] = 0xcafebabe; and watch your program crash and burn. (Sometimes, though, you get an exception you can catch with try{}/catch{}, instead of an actual segv. It would be hilarious... if only the app I was working on did not use this behavior as part of its string-parsing routine...)
              • tptacek 17y ago
                Managed C++ isn't really designed to bulletproof C++; it's designed to make it possible to field C++ code without allowing attackers to upload their own programs into your program's memory. Kicking Managed C++ for not being C# is a bit of a cheap shot.
            • jacquesm 17y ago
              Why is it a strange question ?
          • jrockway 17y ago
            I guess that is a Microsoft term, but basically, if you can't cast an integer to a pointer, you have managed code, and you have eliminated many classes of very severe security vulnerabilities. (I assume, of course, that code like "foo[42]" is logically two unsafe typecasts; pointer to integer, some integer math, and then integer to pointer. The third operation is the security problem.) I don't think managed code is vulnerable to any additional types of exploits, other than holes in the runtime. But remember, if there is a hole in the runtime, you patch the runtime and Every Program Ever Written gets that fix. Everyone knows how to avoid buffer overflows in C, but sometime they mess something up and the app becomes vulnerable. The only way to fix that is to find the problem in the app and fix it... and hope that you got all of them this time. There are plenty of ways managed code can be insecure, of course. I am trying to get root access on my Archos media player, and it is pretty easy because of their poor programming. There is a bash script running as root that downloads a named file from the Internet and puts it in a directory. Except oops, I control the Internet connection and the content that script downloads. I also control the filesystem that it writes to. So while I don't have root, I can get the system to overwrite any file I want, including /etc/shadow, and now I do have root. (Except, the filesystem is read-only, so this doesn't actually work. They got very very lucky. But their code is horribly crappy C, so it should be easy to exploit when I feel like spending time on it.)
    • jergal 17y ago
      :)
    • sown 17y ago
      I was just trying to be helpful.
  • jergal 17y ago
    Amongst other things, for users using gcc ( >= 4.0 ) you have the flag `f-stack-protector' set by default, by setting the `-fno-stack-protector' flag explicitly, you effectively turn of compiler stack canary protection. For those with a non-executable stack, by setting /proc/sys/kernel/randomize_va_space to 0, you effectively turn off any Address Space Layout Randomization (ASLR). If by now, you're still unable to acquire a segmentation violation, try looking into the execshield stack markings, and so on, which may probably very easily be identified via some search engine. Otherwise, it should be trivially capable to overflow a stack based buffer via the conventional routines, should they lack boundary checks. Stack based buffer overflows pose as big a problem now, as they did 13 years ago, and probably more-so.
    • swolchok 17y ago
      The flags for disabling executable stack are: "-Wl,-z execstack" "-Wa,--execstack" I also added -U_FORTIFY_SOURCE in the Makefile for the exploit project that the security course I'm TAing is currently working on. Not sure if that was entirely necessary.
      • jergal 17y ago
        Look at the output of `objdump -d prog | grep call | grep _chk | wc -l', if the value returned is zero/0/0x0 you haven't got FORTIFY_SOURCE enabled, and would be unnecessary to add -U_FORTIFY_SOURCE. However this is dependant upon implementation set restrictions and defaults.