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this post was submitted on 27 Jul 2024
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Programming
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Thanks! The reason I was looking for an example is because I understand:
In principle, but not in practice. Especially the last part.
I have my char buf[16] and some char * ptr = buf; and then a gets() gets a 20 char string, causing a buffer overflow either then or when the buffer is read where it reads out of bounds.
I've done this many times, sometimes intentionally, and if I visualize the memory as one continuous line where the ptr is stored at the precise address buf[20] is at, allowing me to write into that memory location a new address for the pointer by having part of the string given to gets() be a new memory address at the address of ptr, so that next time that pointer is accessed in a program, it leads to an arbitrary memory read, and the arbitrary pointer address can be to still further down in the initial string we gave to gets(), e.g. buf[40] where our shellcode is stored, but how to do this - implement it in practice (so - in code), I don't really know.
Specifically I don't know how to make a pointer at a predictable constant address so it's stored address can be overwritten, and how to make the reading of the resulting maliciously modified pointer also somehow execute code. I'm guessing it can't just be a char pointer reading in data, right?
@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.