4 ms·

IMHO you're converting your heap buffer overflows into stack buffer overflows which are even easier to exploit.

• thenewwazoo 10y ago
  Stack usage is also much, much, much easier to characterize. In systems where stack depth is well-controlled (i.e. most embedded systems that don't have dynamic process/thread creation), very simple analysis will suffice to identify places where you blow your stack.
• usefulcat 10y ago
  Not using the heap != everything is allocated on the stack. In situations where you want to avoid dynamic allocation, memory for most things that would otherwise have been dynamically allocated ends up being statically allocated at compile time.
  • sigjuice 10y ago
    Absolutely none of this is immune to buffer overflows.
    • usefulcat 10y ago
      No, but you're a lot more likely to be able to overwrite the return address via a stack-based buffer overflow, and that is generally a much more serious kind of attack.
• revelation 10y ago
  Exploiting systems without dynamic memory is pretty meh.. that's some NSA level Stuxnet bespoke shit. But no, judging from the code, you just give it one big fat I/O buffer that will usually come from .bss
• nwmcsween 10y ago
  It depends, does malloc have some form of hardening? does the compiler insert stack canaries?