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Diffstat (limited to 'Documentation/locking/lockdep-design.rst')
| -rw-r--r-- | Documentation/locking/lockdep-design.rst | 281 |
1 files changed, 275 insertions, 6 deletions
diff --git a/Documentation/locking/lockdep-design.rst b/Documentation/locking/lockdep-design.rst index 23fcbc4d3fc0..56b90eea2731 100644 --- a/Documentation/locking/lockdep-design.rst +++ b/Documentation/locking/lockdep-design.rst @@ -29,7 +29,7 @@ the validator will shoot a splat if incorrect. A lock-class's behavior is constructed by its instances collectively: when the first instance of a lock-class is used after bootup the class gets registered, then all (subsequent) instances will be mapped to the -class and hence their usages and dependecies will contribute to those of +class and hence their usages and dependencies will contribute to those of the class. A lock-class does not go away when a lock instance does, but it can be removed if the memory space of the lock class (static or dynamic) is reclaimed, this happens for example when a module is @@ -42,6 +42,7 @@ The validator tracks lock-class usage history and divides the usage into (4 usages * n STATEs + 1) categories: where the 4 usages can be: + - 'ever held in STATE context' - 'ever held as readlock in STATE context' - 'ever held with STATE enabled' @@ -49,10 +50,12 @@ where the 4 usages can be: where the n STATEs are coded in kernel/locking/lockdep_states.h and as of now they include: + - hardirq - softirq where the last 1 category is: + - 'ever used' [ == !unused ] When locking rules are violated, these usage bits are presented in the @@ -96,13 +99,13 @@ exact case is for the lock as of the reporting time. +--------------+-------------+--------------+ | | irq enabled | irq disabled | +--------------+-------------+--------------+ - | ever in irq | ? | - | + | ever in irq | '?' | '-' | +--------------+-------------+--------------+ - | never in irq | + | . | + | never in irq | '+' | '.' | +--------------+-------------+--------------+ The character '-' suggests irq is disabled because if otherwise the -charactor '?' would have been shown instead. Similar deduction can be +character '?' would have been shown instead. Similar deduction can be applied for '+' too. Unused locks (e.g., mutexes) cannot be part of the cause of an error. @@ -216,7 +219,7 @@ looks like this:: BD_MUTEX_PARTITION }; -mutex_lock_nested(&bdev->bd_contains->bd_mutex, BD_MUTEX_PARTITION); + mutex_lock_nested(&bdev->bd_contains->bd_mutex, BD_MUTEX_PARTITION); In this case the locking is done on a bdev object that is known to be a partition. @@ -334,7 +337,7 @@ Troubleshooting: ---------------- The validator tracks a maximum of MAX_LOCKDEP_KEYS number of lock classes. -Exceeding this number will trigger the following lockdep warning: +Exceeding this number will trigger the following lockdep warning:: (DEBUG_LOCKS_WARN_ON(id >= MAX_LOCKDEP_KEYS)) @@ -392,3 +395,269 @@ 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: +--------------------- +The whole of the rest document tries to prove a certain type of cycle is equivalent +to deadlock possibility. + +There are three types of lockers: writers (i.e. exclusive lockers, like +spin_lock() or write_lock()), non-recursive readers (i.e. shared lockers, like +down_read()) and recursive readers (recursive shared lockers, like rcu_read_lock()). +And we use the following notations of those lockers in the rest of the document: + + W or E: stands for writers (exclusive lockers). + r: stands for non-recursive readers. + R: stands for recursive readers. + S: stands for all readers (non-recursive + recursive), as both are shared lockers. + N: stands for writers and non-recursive readers, as both are not recursive. + +Obviously, N is "r or W" and S is "r or R". + +Recursive readers, as their name indicates, are the lockers allowed to acquire +even inside the critical section of another reader of the same lock instance, +in other words, allowing nested read-side critical sections of one lock instance. + +While non-recursive readers will cause a self deadlock if trying to acquire inside +the critical section of another reader of the same lock instance. + +The difference between recursive readers and non-recursive readers is because: +recursive readers get blocked only by a write lock *holder*, while non-recursive +readers could get blocked by a write lock *waiter*. Considering the follow +example:: + + TASK A: TASK B: + + read_lock(X); + write_lock(X); + read_lock_2(X); + +Task A gets the reader (no matter whether recursive or non-recursive) on X via +read_lock() first. And when task B tries to acquire writer on X, it will block +and become a waiter for writer on X. Now if read_lock_2() is recursive readers, +task A will make progress, because writer waiters don't block recursive readers, +and there is no deadlock. However, if read_lock_2() is non-recursive readers, +it will get blocked by writer waiter B, and cause a self deadlock. + +Block conditions on readers/writers of the same lock instance: +-------------------------------------------------------------- +There are simply four block conditions: + +1. Writers block other writers. +2. Readers block writers. +3. Writers block both recursive readers and non-recursive readers. +4. And readers (recursive or not) don't block other recursive readers but + may block non-recursive readers (because of the potential co-existing + writer waiters) + +Block condition matrix, Y means the row blocks the column, and N means otherwise. + + +---+---+---+---+ + | | W | r | R | + +---+---+---+---+ + | W | Y | Y | Y | + +---+---+---+---+ + | r | Y | Y | N | + +---+---+---+---+ + | R | Y | Y | N | + +---+---+---+---+ + + (W: writers, r: non-recursive readers, R: recursive readers) + + +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 cannot 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: +--------------------------------------------- +Lock dependencies record the orders of the acquisitions of a pair of locks, and +because there are 3 types for lockers, there are, in theory, 9 types of lock +dependencies, but we can show that 4 types of lock dependencies are enough for +deadlock detection. + +For each lock dependency:: + + L1 -> L2 + +, which means lockdep has seen L1 held before L2 held in the same context at runtime. +And in deadlock detection, we care whether we could get blocked on L2 with L1 held, +IOW, whether there is a locker L3 that L1 blocks L3 and L2 gets blocked by L3. So +we only care about 1) what L1 blocks and 2) what blocks L2. As a result, we can combine +recursive readers and non-recursive readers for L1 (as they block the same types) and +we can combine writers and non-recursive readers for L2 (as they get blocked by the +same types). + +With the above combination for simplification, there are 4 types of dependency edges +in the lockdep graph: + +1) -(ER)->: + exclusive writer to recursive reader dependency, "X -(ER)-> Y" means + X -> Y and X is a writer and Y is a recursive reader. + +2) -(EN)->: + exclusive writer to non-recursive locker dependency, "X -(EN)-> Y" means + X -> Y and X is a writer and Y is either a writer or non-recursive reader. + +3) -(SR)->: + shared reader to recursive reader dependency, "X -(SR)-> Y" means + X -> Y and X is a reader (recursive or not) and Y is a recursive reader. + +4) -(SN)->: + shared reader to non-recursive locker dependency, "X -(SN)-> Y" means + X -> Y and X is a reader (recursive or not) and Y is either a writer or + non-recursive reader. + +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 -(SN)-> Y and X -(EN)-> Y in the dependency graph. + +We use -(xN)-> to represent edges that are either -(EN)-> or -(SN)->, the +similar for -(Ex)->, -(xR)-> and -(Sx)-> + +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 +-(xR)-> and -(Sx)->. In other words, a "strong" path is a path from a lock +walking to another through the lock dependencies, and if X -> Y -> Z is in the +path (where X, Y, Z are locks), and the walk from X to Y is through a -(SR)-> or +-(ER)-> dependency, the walk from Y to Z must not be through a -(SN)-> or +-(SR)-> dependency. + +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 circle), then there is a +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 circle), then there is no +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 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 -(xR)-> and -(Sx)-> (the definition of strong), which +means either L2 in L1 -> L2 is a non-recursive locker (blocked by anyone) or +the L2 in L2 -> L3, is writer (blocking anyone), therefore the holder of L1 +cannot 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 prove that Lx's holder has to wait for +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 it's impossible that Lx on Px+1 is a reader and Lx on Px is a recursive +reader, because readers (no matter recursive or not) don't block recursive +readers, therefore Lx-1 -> Lx and Lx -> Lx+1 cannot be a -(xR)-> -(Sx)-> 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 |