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?