Re: [PATCH] sha: prevent removal of memset as dead store in sha1_update()

[Mikael Pettersson <mikpe@it.uu.se>]

Andi Kleen writes:
 > roel kluin <roel.kluin@gmail.com> writes:
 > 
 > >> And it's wrong because the reason the memset() is there seems to be
 > >> to clear out key information that might exist kernel stack so that
 > >> it's more difficult for rogue code to get at things.
 > >
 > > If the memset is optimized away then the clear out does not occur. Do you
 > > know a different way to fix this? I observed this with:
 > 
 > You could always cast to volatile before memsetting?

I tried that and it doesn't work. Furthermore passing a volatile void *
to a function expecting a void * provokes a compiler warning.

I currently think that defining and using

void secure_bzero(void *p, size_t n)
{
	memset(p, 0, n);
	/* We need for this memset() to be performed even if *p
	 * is about to disappear (a local auto variable going out
	 * of scope or some dynamic memory being kfreed()).
	 * Thus we need to fake a "use" of *p here.
	 * barrier() achieves that effect, and much more.
	 * TODO: find a better alternative to barrier() here.
	 */
	barrier();
}

would be a first good step. We can then ask the gcc folks for
a weaker alternative to barrier() that's guaranteed to keep the
object at [p, p+n[ live.