The compiler knows what memset does. It also knows that stack variables have no use after the function returns. Therefore, the compiler knows there is no reason to write zeroes to this memory, because the program will never read those zeroes. Hence, the compiler will delete the call to memset.
Oh I get it. You're saying the key is stored on the stack and then you can find it by inspecting memory if it hasn't been zeroed out. That's really interesting, what a great example of leaning on language implementation. I guess the correct way to write this is to malloc and free the key. Except, couldn't an attacker see the key anyway while it was live in memory, either on the stack or on the heap (or if it's not "heap", whatever you call the thing that malloc takes memory from)?
I'm not a specialist, but wouldn't a call to free simply deference the memory, but not zero it out? you can then probably still find the key by inspecting memory? again, I'm not a good C programmer, I would like to know too :)
Yeah, you'd still need to call memset before free. Compilers have a much harder time convincing themselves of things about pointers, so it should survive dead code elimination.
