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lockdep/Documention: Recursive read lock detection reasoning
This patch add the documentation piece for the reasoning of deadlock detection related to recursive read lock. The following sections are added: * Explain what is a recursive read lock, and what deadlock cases they could introduce. * Introduce the notations for different types of dependencies, and the definition of strong paths. * Proof for a closed strong path is both sufficient and necessary for deadlock detections with recursive read locks involved. The proof could also explain why we call the path "strong" Signed-off-by: Boqun Feng <boqun.feng@gmail.com> Signed-off-by: Peter Zijlstra (Intel) <peterz@infradead.org> Link: https://lkml.kernel.org/r/20200807074238.1632519-3-boqun.feng@gmail.com
This commit is contained in:
Boqun Feng
Peter Zijlstra
parent
e918188611
commit
224ec489d3
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@ -392,3 +392,261 @@ Run the command and save the output, then compare against the output from
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a later run of this command to identify the leakers. This same output
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can also help you find situations where runtime lock initialization has
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been omitted.
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Recursive read locks:
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---------------------
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The whole of the rest document tries to prove a certain type of cycle is equivalent
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to deadlock possibility.
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There are three types of lockers: writers (i.e. exclusive lockers, like
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spin_lock() or write_lock()), non-recursive readers (i.e. shared lockers, like
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down_read()) and recursive readers (recursive shared lockers, like rcu_read_lock()).
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And we use the following notations of those lockers in the rest of the document:
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W or E: stands for writers (exclusive lockers).
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r: stands for non-recursive readers.
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R: stands for recursive readers.
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S: stands for all readers (non-recursive + recursive), as both are shared lockers.
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N: stands for writers and non-recursive readers, as both are not recursive.
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Obviously, N is "r or W" and S is "r or R".
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Recursive readers, as their name indicates, are the lockers allowed to acquire
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even inside the critical section of another reader of the same lock instance,
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in other words, allowing nested read-side critical sections of one lock instance.
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While non-recursive readers will cause a self deadlock if trying to acquire inside
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the critical section of another reader of the same lock instance.
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The difference between recursive readers and non-recursive readers is because:
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recursive readers get blocked only by a write lock *holder*, while non-recursive
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readers could get blocked by a write lock *waiter*. Considering the follow example:
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TASK A: TASK B:
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read_lock(X);
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write_lock(X);
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read_lock_2(X);
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Task A gets the reader (no matter whether recursive or non-recursive) on X via
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read_lock() first. And when task B tries to acquire writer on X, it will block
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and become a waiter for writer on X. Now if read_lock_2() is recursive readers,
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task A will make progress, because writer waiters don't block recursive readers,
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and there is no deadlock. However, if read_lock_2() is non-recursive readers,
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it will get blocked by writer waiter B, and cause a self deadlock.
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Block conditions on readers/writers of the same lock instance:
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--------------------------------------------------------------
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There are simply four block conditions:
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1. Writers block other writers.
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2. Readers block writers.
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3. Writers block both recursive readers and non-recursive readers.
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4. And readers (recursive or not) don't block other recursive readers but
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may block non-recursive readers (because of the potential co-existing
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writer waiters)
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Block condition matrix, Y means the row blocks the column, and N means otherwise.
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| E | r | R |
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+---+---+---+---+
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E | Y | Y | Y |
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+---+---+---+---+
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r | Y | Y | N |
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+---+---+---+---+
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R | Y | Y | N |
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(W: writers, r: non-recursive readers, R: recursive readers)
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acquired recursively. Unlike non-recursive read locks, recursive read locks
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only get blocked by current write lock *holders* other than write lock
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*waiters*, for example:
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TASK A: TASK B:
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read_lock(X);
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write_lock(X);
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read_lock(X);
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is not a deadlock for recursive read locks, as while the task B is waiting for
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the lock X, the second read_lock() doesn't need to wait because it's a recursive
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read lock. However if the read_lock() is non-recursive read lock, then the above
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case is a deadlock, because even if the write_lock() in TASK B cannot get the
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lock, but it can block the second read_lock() in TASK A.
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Note that a lock can be a write lock (exclusive lock), a non-recursive read
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lock (non-recursive shared lock) or a recursive read lock (recursive shared
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lock), depending on the lock operations used to acquire it (more specifically,
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the value of the 'read' parameter for lock_acquire()). In other words, a single
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lock instance has three types of acquisition depending on the acquisition
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functions: exclusive, non-recursive read, and recursive read.
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To be concise, we call that write locks and non-recursive read locks as
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"non-recursive" locks and recursive read locks as "recursive" locks.
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Recursive locks don't block each other, while non-recursive locks do (this is
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even true for two non-recursive read locks). A non-recursive lock can block the
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corresponding recursive lock, and vice versa.
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A deadlock case with recursive locks involved is as follow:
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TASK A: TASK B:
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read_lock(X);
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read_lock(Y);
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write_lock(Y);
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write_lock(X);
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Task A is waiting for task B to read_unlock() Y and task B is waiting for task
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A to read_unlock() X.
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Dependency types and strong dependency paths:
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---------------------------------------------
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Lock dependencies record the orders of the acquisitions of a pair of locks, and
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because there are 3 types for lockers, there are, in theory, 9 types of lock
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dependencies, but we can show that 4 types of lock dependencies are enough for
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deadlock detection.
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For each lock dependency:
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L1 -> L2
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, which means lockdep has seen L1 held before L2 held in the same context at runtime.
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And in deadlock detection, we care whether we could get blocked on L2 with L1 held,
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IOW, whether there is a locker L3 that L1 blocks L3 and L2 gets blocked by L3. So
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we only care about 1) what L1 blocks and 2) what blocks L2. As a result, we can combine
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recursive readers and non-recursive readers for L1 (as they block the same types) and
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we can combine writers and non-recursive readers for L2 (as they get blocked by the
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same types).
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With the above combination for simplification, there are 4 types of dependency edges
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in the lockdep graph:
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1) -(ER)->: exclusive writer to recursive reader dependency, "X -(ER)-> Y" means
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X -> Y and X is a writer and Y is a recursive reader.
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2) -(EN)->: exclusive writer to non-recursive locker dependency, "X -(EN)-> Y" means
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X -> Y and X is a writer and Y is either a writer or non-recursive reader.
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3) -(SR)->: shared reader to recursive reader dependency, "X -(SR)-> Y" means
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X -> Y and X is a reader (recursive or not) and Y is a recursive reader.
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4) -(SN)->: shared reader to non-recursive locker dependency, "X -(SN)-> Y" means
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X -> Y and X is a reader (recursive or not) and Y is either a writer or
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non-recursive reader.
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Note that given two locks, they may have multiple dependencies between them, for example:
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TASK A:
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read_lock(X);
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write_lock(Y);
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...
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TASK B:
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write_lock(X);
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write_lock(Y);
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, we have both X -(SN)-> Y and X -(EN)-> Y in the dependency graph.
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We use -(xN)-> to represent edges that are either -(EN)-> or -(SN)->, the
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similar for -(Ex)->, -(xR)-> and -(Sx)->
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A "path" is a series of conjunct dependency edges in the graph. And we define a
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"strong" path, which indicates the strong dependency throughout each dependency
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in the path, as the path that doesn't have two conjunct edges (dependencies) as
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-(xR)-> and -(Sx)->. In other words, a "strong" path is a path from a lock
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walking to another through the lock dependencies, and if X -> Y -> Z is in the
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path (where X, Y, Z are locks), and the walk from X to Y is through a -(SR)-> or
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-(ER)-> dependency, the walk from Y to Z must not be through a -(SN)-> or
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-(SR)-> dependency.
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We will see why the path is called "strong" in next section.
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Recursive Read Deadlock Detection:
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----------------------------------
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We now prove two things:
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Lemma 1:
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If there is a closed strong path (i.e. a strong circle), then there is a
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combination of locking sequences that causes deadlock. I.e. a strong circle is
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sufficient for deadlock detection.
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Lemma 2:
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If there is no closed strong path (i.e. strong circle), then there is no
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combination of locking sequences that could cause deadlock. I.e. strong
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circles are necessary for deadlock detection.
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With these two Lemmas, we can easily say a closed strong path is both sufficient
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and necessary for deadlocks, therefore a closed strong path is equivalent to
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deadlock possibility. As a closed strong path stands for a dependency chain that
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could cause deadlocks, so we call it "strong", considering there are dependency
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circles that won't cause deadlocks.
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Proof for sufficiency (Lemma 1):
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Let's say we have a strong circle:
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L1 -> L2 ... -> Ln -> L1
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, which means we have dependencies:
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L1 -> L2
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L2 -> L3
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...
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Ln-1 -> Ln
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Ln -> L1
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We now can construct a combination of locking sequences that cause deadlock:
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Firstly let's make one CPU/task get the L1 in L1 -> L2, and then another get
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the L2 in L2 -> L3, and so on. After this, all of the Lx in Lx -> Lx+1 are
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held by different CPU/tasks.
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And then because we have L1 -> L2, so the holder of L1 is going to acquire L2
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in L1 -> L2, however since L2 is already held by another CPU/task, plus L1 ->
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L2 and L2 -> L3 are not -(xR)-> and -(Sx)-> (the definition of strong), which
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means either L2 in L1 -> L2 is a non-recursive locker (blocked by anyone) or
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the L2 in L2 -> L3, is writer (blocking anyone), therefore the holder of L1
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cannot get L2, it has to wait L2's holder to release.
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Moreover, we can have a similar conclusion for L2's holder: it has to wait L3's
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holder to release, and so on. We now can prove that Lx's holder has to wait for
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Lx+1's holder to release, and note that Ln+1 is L1, so we have a circular
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waiting scenario and nobody can get progress, therefore a deadlock.
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Proof for necessary (Lemma 2):
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Lemma 2 is equivalent to: If there is a deadlock scenario, then there must be a
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strong circle in the dependency graph.
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According to Wikipedia[1], if there is a deadlock, then there must be a circular
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waiting scenario, means there are N CPU/tasks, where CPU/task P1 is waiting for
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a lock held by P2, and P2 is waiting for a lock held by P3, ... and Pn is waiting
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for a lock held by P1. Let's name the lock Px is waiting as Lx, so since P1 is waiting
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for L1 and holding Ln, so we will have Ln -> L1 in the dependency graph. Similarly,
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we have L1 -> L2, L2 -> L3, ..., Ln-1 -> Ln in the dependency graph, which means we
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have a circle:
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Ln -> L1 -> L2 -> ... -> Ln
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, and now let's prove the circle is strong:
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For a lock Lx, Px contributes the dependency Lx-1 -> Lx and Px+1 contributes
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the dependency Lx -> Lx+1, and since Px is waiting for Px+1 to release Lx,
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so it's impossible that Lx on Px+1 is a reader and Lx on Px is a recursive
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reader, because readers (no matter recursive or not) don't block recursive
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readers, therefore Lx-1 -> Lx and Lx -> Lx+1 cannot be a -(xR)-> -(Sx)-> pair,
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and this is true for any lock in the circle, therefore, the circle is strong.
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References:
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-----------
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[1]: https://en.wikipedia.org/wiki/Deadlock
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[2]: Shibu, K. (2009). Intro To Embedded Systems (1st ed.). Tata McGraw-Hill
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