On Sat, Aug 28, 2021 at 03:46:50PM +0800, Sandy Harris wrote: > On Fri, Aug 27, 2021 at 4:40 PM Herbert Xu <herbert@xxxxxxxxxxxxxxxxxxx> wrote: > > > > On Sun, Aug 22, 2021 at 12:31:07PM +0200, Lukas Bulwahn wrote: > > > The function sha512_transform() assigns all local variables to 0 before > > > returning to its caller with the intent to erase sensitive data. > > > .... > > > > > > The assignments to clear a through h and t1/t2 are optimized out by the > > > compiler because they are unused after the assignments. > > Just no. > > You are right, there is a problem here. I thank you for pointing it > out & I've already fixed it in some of my own code. > > However, I think your solution is dead wrong. You are correct that > these assignments are useless because the compiler will optimise them > out, and that's a problem. However, it is not at all "mistiifying"; > they are there for an obvious reason, to avoid leaving state that > might be useful to an enemy. That is quite a small risk, but then it > is a small mitigation, so worth doing. > > The correct solution is not to just remove the assignments, but rather > to replace them with code that will not be optimised away, force the > compiler to do what we need. We already do that for operations that > clear various arrays and structures, using memzero_explicit() rather > than memset(). Similarly, we should replace the assignments with calls > to this macro: > > /* > clear a variable > in a way the compiler will not optimise out > */ > #define clear(x) memzero_explicit( &x, sizeof(x) ) Clearing of local variables is never guaranteed to work properly, as the compiler can create multiple copies and/or put them in registers. It's much more likely to work for arrays than simple variables though (and not cause the variable to be unnecessarily spilled from registers to the stack), so that is the only one the kernel really bothers with. - Eric
