Andi Kleen writes:
> roel kluin <roel.kluin@xxxxxxxxx> 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.
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