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SubjectRe: Re: Re: Re: [RFC][PATCH 2/5] mips/atomic: Fix loongson_llsc_mb() wreckage
On Thu, Apr 25, 2019 at 08:51:17PM +0800, huangpei@loongson.cn wrote:

> > So basically the initial value of @v is set to 1.
> >
> > Then CPU-1 does atomic_add_unless(v, 1, 0)
> > CPU-2 does atomic_set(v, 0)
> >
> > If CPU1 goes first, it will see 1, which is not 0 and thus add 1 to 1
> > and obtains 2. Then CPU2 goes and writes 0, so the exist clause sees
> > v==0 and doesn't observe 2.
> >
> > The other way around, CPU-2 goes first, writes a 0, then CPU-1 goes and
> > observes the 0, finds it matches 0 and doesn't add. Again, the exist
> > clause will find 0 doesn't match 2.
> >
> > This all goes unstuck if interleaved like:
> >
> >
> > CPU-1 CPU-2
> >
> > xor t0, t0
> > 1: ll t0, v
> > bez t0, 2f
> > sw t0, v
> > add t0, t1
> > sc t0, v
> > beqz t0, 1b
> >
> > (sorry if I got the MIPS asm wrong; it's not something I normally write)
> >
> > And the store-word from CPU-2 doesn't make the SC from CPU-1 fail.
> >
>
> loongson's llsc bug DOES NOT fail this litmus( we will not get V=2);
>
> only speculative memory access from CPU-1 can "blind" CPU-1(here blind means do ll/sc
> wrong), this speculative memory access can be observed corrently by CPU2. In this
> case, sw from CPU-2 can get I , which can be observed by CPU-1, and clear llbit,then
> failed sc.

I'm not following, suppose CPU-1 happens as a speculation (imagine
whatever code is required to make that happen before). CPU-2 sw will
cause I on CPU-1's ll but, as in the previous email, CPU-1 will continue
as if it still has E and complete the SC.

That is; I'm just not seeing why this case would be different from two
competing LL/SCs.

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