[PATCH v3 1/7] dump_stack: Support adding to the dump stack arch description

Petr Mladek pmladek at suse.com
Thu Feb 21 19:38:35 AEDT 2019


On Wed 2019-02-20 14:44:33, Andrea Parri wrote:
> > >> > > +		 * Order the stores above in vsnprintf() vs the store of the
> > >> > > +		 * space below which joins the two strings. Note this doesn't
> > >> > > +		 * make the code truly race free because there is no barrier on
> > >> > > +		 * the read side. ie. Another CPU might load the uninitialised
> > >> > > +		 * tail of the buffer first and then the space below (rather
> > >> > > +		 * than the NULL that was there previously), and so print the
> > >> > > +		 * uninitialised tail. But the whole string lives in BSS so in
> > >> > > +		 * practice it should just see NULLs.
> > >> > 
> > It is not my intention to support concurrent updates of the string. The
> > idea is you setup the string early in boot.
> 
> Understood, thanks for the clarification.
> > 
> > The concern with a concurrent reader is simply that the string is dumped
> > in the panic path, and you never really know when you're going to panic.
> > Even if you only write to the string before doing SMP bringup you might
> > still have another CPU go rogue and panic before then.
> > 
> > But I probably should have just not added the barrier, it's over
> > paranoid and will almost certainly never matter in practice.
> 
> Oh, well, I can only echo you: if you don't care about the stores being
> _observed_ out of order, you could simply remove the barrier; if you do
> care, then you need "more paranoid" on the readers side.  ;-)

Hmm, the barrier might be fine and actually useful. The
purpose is to make sure that the later '\0' is written before
the existing one is replaced by ' '.

The reader does not need the barrier as long as it reads the string
sequentially. I would expect that it is always the case. But who
knows with all the speculation-related CPU bugs around.

In each case, any race could never crash the kernel.
The dump_stack_arch_desc_str is zeroed out of box and
the very last '\0' is never rewritten.

Best Regards,
Petr


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