On Fri, Feb 23, 2018 at 12:55:20PM +0100, Peter Zijlstra wrote:
> On Thu, Feb 22, 2018 at 03:08:51PM +0800, Boqun Feng wrote:
> > @@ -1012,6 +1013,33 @@ static inline bool bfs_error(enum bfs_result res)
> > return res < 0;
> > }
> >
> > +#define DEP_NN_BIT 0
> > +#define DEP_RN_BIT 1
> > +#define DEP_NR_BIT 2
> > +#define DEP_RR_BIT 3
> > +
> > +#define DEP_NN_MASK (1U << (DEP_NN_BIT))
> > +#define DEP_RN_MASK (1U << (DEP_RN_BIT))
> > +#define DEP_NR_MASK (1U << (DEP_NR_BIT))
> > +#define DEP_RR_MASK (1U << (DEP_RR_BIT))
> > +
> > +static inline unsigned int __calc_dep_bit(int prev, int next)
> > +{
> > + if (prev == 2 && next != 2)
> > + return DEP_RN_BIT;
> > + if (prev != 2 && next == 2)
> > + return DEP_NR_BIT;
> > + if (prev == 2 && next == 2)
> > + return DEP_RR_BIT;
> > + else
> > + return DEP_NN_BIT;
> > +}
> > +
> > +static inline unsigned int calc_dep(int prev, int next)
> > +{
> > + return 1U << __calc_dep_bit(prev, next);
> > +}
> > +
> > static enum bfs_result __bfs(struct lock_list *source_entry,
> > void *data,
> > int (*match)(struct lock_list *entry, void *data),
> > @@ -1921,6 +1949,16 @@ check_prev_add(struct task_struct *curr, struct
> > held_lock *prev,
> > if (entry->class == hlock_class(next)) {
> > if (distance == 1)
> > entry->distance = 1;
> > + entry->dep |= calc_dep(prev->read, next->read);
> > + }
> > + }
> > +
> > + /* Also, update the reverse dependency in @next's ->locks_before list */
> > + list_for_each_entry(entry, &hlock_class(next)->locks_before, entry) {
> > + if (entry->class == hlock_class(prev)) {
> > + if (distance == 1)
> > + entry->distance = 1;
> > + entry->dep |= calc_dep(next->read, prev->read);
> > return 1;
> > }
> > }
>
> I think it all becomes simpler if you use only 2 bits. Such that:
>
> bit0 is the prev R (0) or N (1) value,
> bit1 is the next R (0) or N (1) value.
>
> I think this should work because we don't care about the empty set
> (currently 0000) and all the complexity in patch 5 is because we can
> have R bits set when there's also N bits. The concequence of that is
> that we cannot replace ! with ~ (which is what I kept doing).
>
> But with only 2 bits, we only track the strongest relation in the set,
> which is exactly what we appear to need.
>
But if we only have RN and NR, both bits will be set, we can not check
whether we have NN or not. Consider we have:
A -(RR)-> B
B -(NR)-> C and B -(RN)-> C
C -(RN)-> A
this is not a deadlock case, but with "two bits" approach, we can not
differ this with:
A -(RR)-> B
B -(NN)-> C
C -(RN)-> A
, which is a deadlock.
But maybe "three bits" (NR, RN and NN bits) approach works, that is if
->dep is 0, we indicates this is only RR, and is_rx() becomes:
static inline bool is_rx(u8 dep)
{
return !(dep & (NR_MASK | NN_MASK));
}
and is_xr() becomes:
static inline bool is_xr(u8 dep)
{
return !(dep & (RN_MASK | NN_MASK));
}
, with this I think your simplification with have_xr works, thanks!
Regards,
Boqun
>
> The above then becomes something like:
>
> static inline u8 __calc_dep(struct held_lock *lock)
> {
> return lock->read != 2;
> }
>
> static inline u8
> calc_dep(struct held_lock *prev, struct held_lock *next)
> {
> return (__calc_dep(prev) << 0) | (__calc_dep(next) << 1);
> }
>
>
> entry->dep |= calc_dep(prev, next);
>
>
>
> Then the stuff from 5 can be:
>
> static inline bool is_rx(u8 dep)
> {
> return !(dep & 1);
> }
>
> static inline bool is_xr(u8 dep)
> {
> return !(dep & 2);
> }
>
>
> if (have_xr && is_rx(entry->dep))
> continue;
>
> entry->have_xr = is_xr(entry->dep);
>
>
> Or did I mess that up somewhere?
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