[PATCH v3 0/6] implement KASLR for powerpc/fsl_booke/64

[Scott Wood]

On Fri, 2020-02-28 at 14:47 +0800, Jason Yan wrote:
> 
> 在 2020/2/28 13:53, Scott Wood 写道:
> > On Wed, 2020-02-26 at 16:18 +0800, Jason Yan wrote:
> > > Hi Daniel,
> > > 
> > > 在 2020/2/26 15:16, Daniel Axtens 写道:
> > > > Maybe replacing the REG format string in KASLR mode would be
> > > > sufficient?
> > > > 
> > > 
> > > Most archs have removed the address printing when dumping stack. Do we
> > > really have to print this?
> > > 
> > > If we have to do this, maybe we can use "%pK" so that they will be
> > > hidden from unprivileged users.
> > 
> > I've found the addresses to be useful, especially if I had a way to dump
> > the
> > stack data itself.  Wouldn't the register dump also be likely to give away
> > the
> > addresses?
> 
> If we have to print the address, then kptr_restrict and dmesg_restrict
> must be set properly so that unprivileged users cannot see them.

And how does that work with crash dumps that could be from any context?

dmesg_restrict is irrelevant as it just controls who can see the dmesg, not
what goes into it.  kptr_restrict=1 will only get the value if you're not in
any sort of IRQ, *and* if the crashing context happened to have CAP_SYSLOG. 
No other value of kptr_restrict will ever get you the raw value.

> > 
> > I don't see any debug setting for %pK (or %p) to always print the actual
> > address (closest is kptr_restrict=1 but that only works in certain
> > contexts)... from looking at the code it seems it hashes even if kaslr is
> > entirely disabled?  Or am I missing something?
> > 
> 
> Yes, %pK (or %p) always hashes whether kaslr is disabled or not. So if
> we want the real value of the address, we cannot use it. But if you only
> want to distinguish if two pointers are the same, it's ok.

Am I the only one that finds this a bit crazy?  If you want to lock a system
down then fine, but why wage war on debugging even when there's no
randomization going on?  Comparing two pointers for equality is not always
adequate.

-Scott