Re: [RFC tip/locking/lockdep v5 04/17] lockdep: Introduce lock_list::dep
From: Boqun Feng
Date: Sat Feb 24 2018 - 03:57:02 EST
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.
Am I missing something sublte?
Regards,
Boqun
> So can you elaborate a bit?
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