On Mon, Feb 26, 2018 at 10:00:50AM +0100, Peter Zijlstra wrote:
> On Sat, Feb 24, 2018 at 05:26:52PM +0800, Boqun Feng wrote:
> > > 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?
> 
> Yes, although it took me forever to grasp because I have snot for brains
> atm :-(
> 

Take care!

> Would something like:
> 
> 
> 	dep = entry->dep;
> 
> 	/* Mask out all *R -> R* relations. */
> 	if (have_xr)
> 		dep &= ~(RR_MASK | RN_MASK);
> 
> 	/* If nothing left, we're done. */
> 	if (!dep)
> 		continue;
> 
> 	/* If there are (only) *R left, set that for the next step. */
> 	entry->have_xr = !(dep & (RN_MASK | NN_MASK));
> 
> 
> Work? I think that reads fairly simple.

I think that works, will apply this simplification to see whether the
self tests agree.

Btw, given the comments in the code, I think it's better to name
"have_xr" as "only_xr"? I have a feeling that my comment-less code
somehow misled you to that name ;-(

Regards,
Boqun