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this post was submitted on 27 Jul 2024
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Programming
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@LainTrain The simplest case is overwriting the return address on the stack. If your stack layout looks like this (B for buffer, R for return address, A for function arguments):
BBBBBBBBRRRRAAAA
and you give a pointer to the first B byte to gets(), the input can overwrite the bytes of R.
You can try this with a 32-bit program complied with disabled mitigations. Run the program in a debugger, break in the function, inspect the stack pointer value. With ASLR disabled the addresses will remain the same for every program execution assuming the call graph at this point doesn't change. You can then overwrite the bytes of R with the buffer address (assuming no stack canary), and overwrite the buffer bytes with machine code instructions. When the function attempts to return, it instead jumps to the instructions you left in the buffer, and executes them (assuming no W^X).
Thank you! This is incredibly helpful and insightful.
I now understand how one would do this with manually writing in a debugger, am I correct in thinking that if I constructed the input to gets() in such a manner that BBBBBBB contains shellcode, and RRRR is a return address pointing to the beginning of BBBBB then that is how arbitrary code execution can be achieved with this in practice?
@LainTrain Yes, but "in practice" this simple approach worked 20 years ago. Modern processors, compilers and operating systems make exploitation of stack buffer overflows a lot more difficult.
That's fine, I think for my purposes it's better to start simple with the basic concept of it first, then add complexity by learning about the protections and how they have/could be circumvented.