[PATCH v4 2/2] powerpc/rtas: Implement reentrant rtas call

[Nicholas Piggin]

Excerpts from Leonardo Bras's message of May 15, 2020 9:51 am:
> Implement rtas_call_reentrant() for reentrant rtas-calls:
> "ibm,int-on", "ibm,int-off",ibm,get-xive" and  "ibm,set-xive".
> 
> On LoPAPR Version 1.1 (March 24, 2016), from 7.3.10.1 to 7.3.10.4,
> items 2 and 3 say:
> 
> 2 - For the PowerPC External Interrupt option: The * call must be
> reentrant to the number of processors on the platform.
> 3 - For the PowerPC External Interrupt option: The * argument call
> buffer for each simultaneous call must be physically unique.
> 
> So, these rtas-calls can be called in a lockless way, if using
> a different buffer for each cpu doing such rtas call.

What about rtas_call_unlocked? Do the callers need to take the rtas 
lock?

Machine checks must call ibm,nmi-interlock too, which we really don't 
want to take a lock for either. Hopefully that's in a class of its own
and we can essentially ignore with respect to other rtas calls.

The spec is pretty vague too :(

"The ibm,get-xive call must be reentrant to the number of processors on 
the platform."

This suggests ibm,get-xive can be called concurrently by multiple
processors. It doesn't say anything about being re-entrant against any 
of the other re-entrant calls. Maybe that could be reasonably assumed,
but I don't know if it's reasonable to assume it can be called 
concurrently with a *non-reentrant* call, is it?

> For this, it was suggested to add the buffer (struct rtas_args)
> in the PACA struct, so each cpu can have it's own buffer.

You can't do this, paca is not limited to RTAS_INSTANTIATE_MAX.
Which is good, because I didn't want you to add another 88 bytes to the 
paca :) Can you make it a pointer and allocate it separately? Check
the slb_shadow allocation, you could use a similar pattern.

The other option would be to have just one more rtas args, and have the 
crashing CPU always that. That would skirt the re-entrancy issue -- the
concurrency is only ever a last resort. Would be a bit tricker though.

Thanks,
Nick