Commit 224ec489 authored by Boqun Feng's avatar Boqun Feng Committed by Peter Zijlstra

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: default avatarBoqun Feng <boqun.feng@gmail.com>
Signed-off-by: default avatarPeter Zijlstra (Intel) <peterz@infradead.org>
Link: https://lkml.kernel.org/r/20200807074238.1632519-3-boqun.feng@gmail.com
parent e9181886
......@@ -392,3 +392,261 @@ 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.
| E | r | R |
+---+---+---+---+
E | 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
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