Re: [RFC tip/locking/lockdep v5 04/17] lockdep: Introduce lock_list::dep

[Boqun Feng <boqun.feng@gmail.com>]

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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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