On Sat, Feb 24, 2018 at 05:26:52PM +0800, Boqun Feng wrote:
> > This is for case like:
> >
> > TASK1:
> > read_lock(A);
> > read_lock(B);
> >
> > TASK2:
> > write_lock(B);
> > read_lock(C);
> >
> > TASK3:
> > read_lock(B);
> > write_lock(C);
> >
> > TASK4:
> > read_lock(C);
> > write_lock(A);
> >
> > , which is not a deadlock.
> >
>
> After TASK 1,2,3 have executed, we have A -(RR)-> B, B -(RN/NR)-> C, and
> when TASK4 executed, we will try to add C -(RN)-> A into the graph.
> Before that we need to check whether we have a A -> ... -(*N)-> C path
> in the graph already, so we search from A (@prev is C and @this is A):
>
> * we set A->have_xr to false, because the dependency we are adding
> is a RN.
>
> * we find A -(RR)-> B, and since have_xr (= A->have_xr) is false,
> we can pick this dependency, and since for A -> B, we only have
> RR, so we set B->have_xr to true.
>
> * we then find B -(RN/NR)-> C, and since have_xr (= B->have_xr) is
> true, we will pick it only only_rx(C->dep) return false,
> otherwise we skip. Because we have RN and NR for B -> C,
> therefore we won't skip B -> C.
>
> * Now we try to set C->have_xr, if we set it to only_xr(C->dep),
> we will set it to false, right? Because B -> C has RN.
>
> * Since we now find a entry equal to @prev, we go into the
> hlock_conflict() logic and for expression
>
> hlock->read != 2 || !entry->have_xr
>
> @hlock is the C in TASK4, so hlock->read == 2, and @entry is the
> C whose ->have_xr we just set, so entry->have_xr is false.
> Therefore hlock_conflict() returns true. And that indicates we
> find a deadlock in the search. But the above senario can not
> introduce a deadlock.
>
> Could this help you, or I miss something?
Yes, although it took me forever to grasp because I have snot for brains
atm :-(
Would something like:
dep = entry->dep;
/* Mask out all *R -> R* relations. */
if (have_xr)
dep &= ~(RR_MASK | RN_MASK);
/* If nothing left, we're done. */
if (!dep)
continue;
/* If there are (only) *R left, set that for the next step. */
entry->have_xr = !(dep & (RN_MASK | NN_MASK));
Work? I think that reads fairly simple.
