Re: Internal vs. external barriers (was: Re: Interesting LKMM litmus test)

Paul E. McKenney
2023-01-20T21:37:35Z

On Fri, Jan 20, 2023 at 09:46:55PM +0100, Jonas Oberhauser wrote:
>
>
> On 1/20/2023 4:39 PM, Paul E. McKenney wrote:
> > On Fri, Jan 20, 2023 at 10:43:10AM +0100, Jonas Oberhauser wrote:
> >
> > > I don't think Boqun's patch is hard to repair.
> > > Besides the issue you mention, I think it's also missing Sync-srcu, which
> > > seems to be linked by loc based on its first argument.
> > >
> > > How about something like this?
> > >
> > > let ALL-LOCKS = LKR | LKW | UL | LF | RU | Srcu-lock | Srcu-unlock |
> > > Sync-srcu flag ~empty ~[ALL_LOCKS | IW] ; loc ; [ALL-LOCKS] as
> > > mixed-lock-accesses
> > >
> > > If you're using something that isn't a lock or intial write on the same location as a lock, you get the flag.
> > Wouldn't that unconditionally complain about the first srcu_read_lock()
> > in a given process? Or am I misreading those statements?
> >
>
> I unfolded the definition step by step and it seems I was careless when
> distributing the ~ over the [] operator.
> I should have written:
>
> flag ~empty [~(ALL_LOCKS | IW)] ; loc ; [ALL-LOCKS] as mixed-lock-accesses
>
> but somehow I thought I can save the parentheses by putting the ~ on the
> outside.
> Now on the off-chance that this is kind of how you already read the
> relation, let me unfold it step-by-step.
>
> Let's assume that the sequence s of operations on this location is
>   s = initial write , (perhaps some gps) , first read lock , read
> lock&unlock&gp ...
> then the flag would appear if the specified relation isn't empty. That would
> be the case if there are a and b that are linked by
>
> a ->[~(ALL_LOCKS | IW)] ; loc ; [ALL-LOCKS] b
>
> This means a is neither in ALL_LOCKS nor in IW, while b is ALL-LOCKS; and furthermore, they are equal to a' and b' resp. that are related by loc, i.e., appear in this sequence s. Thus both a and b are actually appearing both in the sequence s.
> However, every event in the sequence s is either in ALL_LOCKS or in IW, which contradicts the assumption that a is in the sequence and in neither of the sets. Because of this contradiction, the flag doesn't appear if the sequence looks like this.
>
> More generally, if every event in the sequence is either the initial write or one of (srcu-) lock,unlock,up,down,sync, there won't be a flag.
>
> In contrast, if the sequence has the form
> s' = initial write, (normal srcu events), some other acces x, (normal srcu events)
> and y is one of the srcu events in this sequence, then
> x ->[~(ALL_LOCKS | IW)] ; loc ; [ALL_LOCKS] y
> and you get a flag.

Thank you! When I get done messing with NMIs, I will give this a go.

Just out of curiosity, are you sent up to run LKMM locally at your end?

Thanx, Paul