Hi, Just a few typos etc. below...

On 04/11/2018 06:50 AM, Boqun Feng wrote: > Signed-off-by: Boqun Feng <boqun.f...@gmail.com> > --- > Documentation/locking/lockdep-design.txt | 178 > +++++++++++++++++++++++++++++++ > 1 file changed, 178 insertions(+) > > diff --git a/Documentation/locking/lockdep-design.txt > b/Documentation/locking/lockdep-design.txt > index 9de1c158d44c..6bb9e90e2c4f 100644 > --- a/Documentation/locking/lockdep-design.txt > +++ b/Documentation/locking/lockdep-design.txt > @@ -284,3 +284,181 @@ Run the command and save the output, then compare > against the output from > a later run of this command to identify the leakers. This same output > can also help you find situations where runtime lock initialization has > been omitted. > + > +Recursive read locks: > +--------------------- > + > +Lockdep now is equipped with deadlock detection for recursive read locks. > + > +Recursive read locks, as their name indicates, are the locks able to be > +acquired recursively. Unlike non-recursive read locks, recursive read locks > +only get blocked by current write lock *holders* other than write lock > +*waiters*, for example: > + > + TASK A: TASK B: > + > + read_lock(X); > + > + write_lock(X); > + > + read_lock(X); > + > +is not a deadlock for recursive read locks, as while the task B is waiting > for > +the lock X, the second read_lock() doesn't need to wait because it's a > recursive > +read lock. However if the read_lock() is non-recursive read lock, then the > above > +case is a deadlock, because even if the write_lock() in TASK B can not get > the > +lock, but it can block the second read_lock() in TASK A. > + > +Note that a lock can be a write lock (exclusive lock), a non-recursive read > +lock (non-recursive shared lock) or a recursive read lock (recursive shared > +lock), depending on the lock operations used to acquire it (more > specifically, > +the value of the 'read' parameter for lock_acquire()). In other words, a > single > +lock instance has three types of acquisition depending on the acquisition > +functions: exclusive, non-recursive read, and recursive read. > + > +To be concise, we call that write locks and non-recursive read locks as > +"non-recursive" locks and recursive read locks as "recursive" locks. > + > +Recursive locks don't block each other, while non-recursive locks do (this is > +even true for two non-recursive read locks). A non-recursive lock can block > the > +corresponding recursive lock, and vice versa. > + > +A deadlock case with recursive locks involved is as follow: > + > + TASK A: TASK B: > + > + read_lock(X); > + read_lock(Y); > + write_lock(Y); > + write_lock(X); > + > +Task A is waiting for task B to read_unlock() Y and task B is waiting for > task > +A to read_unlock() X. > + > +Dependency types and strong dependency paths: > +--------------------------------------------- > +In order to detect deadlocks as above, lockdep needs to track different > dependencies. > +There are 4 categories for dependency edges in the lockdep graph: > + > +1) -(NN)->: non-recursive to non-recursive dependency. "X -(NN)-> Y" means > + X -> Y and both X and Y are non-recursive locks. > + > +2) -(RN)->: recursive to non-recursive dependency. "X -(RN)-> Y" means > + X -> Y and X is recursive read lock and Y is non-recursive lock. > + > +3) -(NR)->: non-recursive to recursive dependency, "X -(NR)-> Y" means > + X -> Y and X is non-recursive lock and Y is recursive lock. > + > +4) -(RR)->: recursive to recursive dependency, "X -(RR)-> Y" means > + X -> Y and both X and Y are recursive locks. > + > +Note that given two locks, they may have multiple dependencies between them, > for example: > + > + TASK A: > + > + read_lock(X); > + write_lock(Y); > + ... > + > + TASK B: > + > + write_lock(X); > + write_lock(Y); > + > +, we have both X -(RN)-> Y and X -(NN)-> Y in the dependency graph. > + > +We use -(*N)-> for edges that is either -(RN)-> or -(NN)->, the similar for > -(N*)->, > +-(*R)-> and -(R*)-> > + > +A "path" is a series of conjunct dependency edges in the graph. And we > define a > +"strong" path, which indicates the strong dependency throughout each > dependency > +in the path, as the path that doesn't have two conjunct edges (dependencies) > as > +-(*R)-> and -(R*)->. In other words, a "strong" path is a path from a lock > +walking to another through the lock dependencies, and if X -> Y -> Z in the > +path (where X, Y, Z are locks), if the walk from X to Y is through a -(NR)-> > or > +-(RR)-> dependency, the walk from Y to Z must not be through a -(RN)-> or > +-(RR)-> dependency, otherwise it's not a strong path. > + > +We will see why the path is called "strong" in next section. > + > +Recursive Read Deadlock Detection: > +---------------------------------- > + > +We now prove two things: > + > +Lemma 1: > + > +If there is a closed strong path (i.e. a strong cirle), then there is a ?? circle > +combination of locking sequences that causes deadlock. I.e. a strong circle > is > +sufficient for deadlock detection. > + > +Lemma 2: > + > +If there is no closed strong path (i.e. strong cirle), then there is no ?? circle > +combination of locking sequences that could cause deadlock. I.e. strong > +circles are necessary for deadlock detection. > + > +With these two Lemmas, we can easily say a closed strong path is both > sufficient > +and necessary for deadlocks, therefore a closed strong path is equivalent to > +deadlock possibility. As a closed strong path stands for a dependency chain > that > +could cause deadlocks, so we call it "strong", considering there are > dependency > +circles that won't cause deadlocks. > + > +Proof for sufficiency (Lemma 1): > + > +Let's say we have a strong cirlce: circle: > + > + L1 -> L2 ... -> Ln -> L1 > + > +, which means we have dependencies: > + > + L1 -> L2 > + L2 -> L3 > + ... > + Ln-1 -> Ln > + Ln -> L1 > + > +We now can construct a combination of locking sequences that cause deadlock: > + > +Firstly let's make one CPU/task get the L1 in L1 -> L2, and then another get > +the L2 in L2 -> L3, and so on. After this, all of the Lx in Lx -> Lx+1 are > +held by different CPU/tasks. > + > +And then because we have L1 -> L2, so the holder of L1 is going to acquire L2 > +in L1 -> L2, however since L2 is already held by another CPU/task, plus L1 -> > +L2 and L2 -> L3 are not *R and R* (the definition of strong), therefore the > +holder of L1 can not get L2, it has to wait L2's holder to release. > + > +Moreover, we can have a similar conclusion for L2's holder: it has to wait > L3's > +holder to release, and so on. We now can proof that Lx's holder has to wait > for prove > +Lx+1's holder to release, and note that Ln+1 is L1, so we have a circular > +waiting scenario and nobody can get progress, therefore a deadlock. > + > +Proof for necessary (Lemma 2): > + > +Lemma 2 is equivalent to: If there is a deadlock scenario, then there must > be a > +strong circle in the dependency graph. > + > +According to Wikipedia[1], if there is a deadlock, then there must be a > circular > +waiting scenario, means there are N CPU/tasks, where CPU/task P1 is waiting > for > +a lock held by P2, and P2 is waiting for a lock held by P3, ... and Pn is > waiting > +for a lock held by P1. Let's name the lock Px is waiting as Lx, so since P1 > is waiting > +for L1 and holding Ln, so we will have Ln -> L1 in the dependency graph. > Similarly, > +we have L1 -> L2, L2 -> L3, ..., Ln-1 -> Ln in the dependency graph, which > means we > +have a circle: > + > + Ln -> L1 -> L2 -> ... -> Ln > + > +, and now let's prove the circle is strong: > + > +For a lock Lx, Px contributes the dependency Lx-1 -> Lx and Px+1 contributes > +the dependency Lx -> Lx+1, and since Px is waiting for Px+1 to release Lx, > +so Lx can not be both recursive in Lx -> Lx+1 and Lx-1 -> Lx, because > recursive > +locks don't block each other, therefore Lx-1 -> Lx and Lx -> Lx+1 can not be > a > +-(*R)-> -(R*)-> pair, and this is true for any lock in the circle, therefore, > +the circle is strong. > + > +References: > +----------- > +[1]: https://en.wikipedia.org/wiki/Deadlock > +[2]: Shibu, K. (2009). Intro To Embedded Systems (1st ed.). Tata McGraw-Hill > I would also change all /can not/ to /cannot/... -- ~Randy