On Sat, Feb 24, 2018 at 05:00:19PM +0800, Boqun Feng wrote:
> On Sat, Feb 24, 2018 at 09:38:07AM +0100, Peter Zijlstra wrote:
> > On Sat, Feb 24, 2018 at 02:30:05PM +0800, Boqun Feng wrote:
> > > On Sat, Feb 24, 2018 at 01:32:50PM +0800, Boqun Feng wrote:
> >
> > > > /*
> > > > * DEP_*_BIT in lock_list::dep
> > > > *
> > > > * For dependency @prev -> @next:
> > > > *
> > > > * RR: both @prev and @next are recursive read locks, i.e.
> > > > ->read == 2.
> > > > * RN: @prev is recursive and @next is non-recursive.
> > > > * NR: @prev is a not recursive and @next is recursive.
> > > > * NN: both @prev and @next are non-recursive.
> > > > *
> > > > * Note that we define the value of DEP_*_BITs so that:
> > > > * bit0 is prev->read != 2
> > > > * bit1 is next->read != 2
> > > > */
> > > > #define DEP_RR_BIT 0
> > > > #define DEP_RN_BIT 1
> > > > #define DEP_NR_BIT 2
> > > > #define DEP_NN_BIT 3
> > > >
> > > > #define DEP_RR_MASK (1U << (DEP_RR_BIT))
> > > > #define DEP_RN_MASK (1U << (DEP_RN_BIT))
> > > > #define DEP_NR_MASK (1U << (DEP_NR_BIT))
> > > > #define DEP_NN_MASK (1U << (DEP_NN_BIT))
> > > >
> > > > static inline unsigned int
> > > > __calc_dep_bit(struct held_lock *prev, struct held_lock *next)
> > > > {
> > > > return (prev->read != 2) + ((next->read != 2) << 1)
> > > > }
> > > >
> > > > static inline u8 calc_dep(struct held_lock *prev, struct
> > > > held_lock *next)
> > > > {
> > > > return 1U << __calc_dep_bit(prev, next);
> > > > }
> > > >
> > > > static inline bool only_rx(u8 dep)
> > > > {
> > > > return !(dep & (DEP_NR_MASK | DEP_NN_MASK));
> > > > }
> > > >
> > > > static inline bool only_xr(u8 dep)
> > > > {
> > > > return !(dep & (DEP_NR_MASK | DEP_NN_MASK));
> > > > }
> > > >
> >
> > > > > > if (have_xr && is_rx(entry->dep))
> > > > > > continue;
> > > > > >
> > > > > > entry->have_xr = is_xr(entry->dep);
> > > > > >
> > >
> > > Hmm.. I think this part also needs some tweak:
> > >
> > > /* if -> prev is *R, and we only have R* for prev -> this, * skip*/
> > > if (have_xr && only_rx(entry->dep))
> > > continue;
> > >
> > > /*
> > > * we pick a *R for prev -> this only if:
> > > * prev -> this dependencies are all *R
> > > * or
> > > * -> prev is *R, and we don't have NN for prev -> this
> > > */
> > > entry->have_xr = only_xr(entry->dep) || (have_xr && !is_nn(entry->dep));
> > >
> > > otherwise, we will wrongly set entry->have_xr to false if have_xr is
> > > true and we have RN for prev -> this.
> >
> > OK, so its saturday morning and such, but what? Why should we set
> > have_xr true when we have RN? Note that if we only had RN we'd already
> > have bailed on the continue due to only_rx().
> >
>
> But what if we have RN and NR? only_rx() will return false, but due to
> have_xr is true, we can not pick RN, so entry->have_xr should be set to
> true (due to we have to pick NR), however only_xr() is false becuase we
> have RN, so if we set entry->have_xr to only_xr(), we set it as false.
>
> 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?
Regards,
Boqun
> Am I missing something sublte?
>
>
> Regards,
> Boqun
>
> > So can you elaborate a bit?
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