Commit 25595eb6 authored by Mathieu Desnoyers's avatar Mathieu Desnoyers Committed by Peter Zijlstra

sched: membarrier: document memory ordering scenarios

Document membarrier ordering scenarios in membarrier.c. Thanks to Alan
Stern for refreshing my memory. Now that I have those in mind, it seems
appropriate to serialize them to comments for posterity.
Signed-off-by: default avatarMathieu Desnoyers <mathieu.desnoyers@efficios.com>
Signed-off-by: default avatarPeter Zijlstra (Intel) <peterz@infradead.org>
Link: https://lkml.kernel.org/r/20201020134715.13909-4-mathieu.desnoyers@efficios.com
parent 618758ed
...@@ -6,6 +6,134 @@ ...@@ -6,6 +6,134 @@
*/ */
#include "sched.h" #include "sched.h"
/*
* For documentation purposes, here are some membarrier ordering
* scenarios to keep in mind:
*
* A) Userspace thread execution after IPI vs membarrier's memory
* barrier before sending the IPI
*
* Userspace variables:
*
* int x = 0, y = 0;
*
* The memory barrier at the start of membarrier() on CPU0 is necessary in
* order to enforce the guarantee that any writes occurring on CPU0 before
* the membarrier() is executed will be visible to any code executing on
* CPU1 after the IPI-induced memory barrier:
*
* CPU0 CPU1
*
* x = 1
* membarrier():
* a: smp_mb()
* b: send IPI IPI-induced mb
* c: smp_mb()
* r2 = y
* y = 1
* barrier()
* r1 = x
*
* BUG_ON(r1 == 0 && r2 == 0)
*
* The write to y and load from x by CPU1 are unordered by the hardware,
* so it's possible to have "r1 = x" reordered before "y = 1" at any
* point after (b). If the memory barrier at (a) is omitted, then "x = 1"
* can be reordered after (a) (although not after (c)), so we get r1 == 0
* and r2 == 0. This violates the guarantee that membarrier() is
* supposed by provide.
*
* The timing of the memory barrier at (a) has to ensure that it executes
* before the IPI-induced memory barrier on CPU1.
*
* B) Userspace thread execution before IPI vs membarrier's memory
* barrier after completing the IPI
*
* Userspace variables:
*
* int x = 0, y = 0;
*
* The memory barrier at the end of membarrier() on CPU0 is necessary in
* order to enforce the guarantee that any writes occurring on CPU1 before
* the membarrier() is executed will be visible to any code executing on
* CPU0 after the membarrier():
*
* CPU0 CPU1
*
* x = 1
* barrier()
* y = 1
* r2 = y
* membarrier():
* a: smp_mb()
* b: send IPI IPI-induced mb
* c: smp_mb()
* r1 = x
* BUG_ON(r1 == 0 && r2 == 1)
*
* The writes to x and y are unordered by the hardware, so it's possible to
* have "r2 = 1" even though the write to x doesn't execute until (b). If
* the memory barrier at (c) is omitted then "r1 = x" can be reordered
* before (b) (although not before (a)), so we get "r1 = 0". This violates
* the guarantee that membarrier() is supposed to provide.
*
* The timing of the memory barrier at (c) has to ensure that it executes
* after the IPI-induced memory barrier on CPU1.
*
* C) Scheduling userspace thread -> kthread -> userspace thread vs membarrier
*
* CPU0 CPU1
*
* membarrier():
* a: smp_mb()
* d: switch to kthread (includes mb)
* b: read rq->curr->mm == NULL
* e: switch to user (includes mb)
* c: smp_mb()
*
* Using the scenario from (A), we can show that (a) needs to be paired
* with (e). Using the scenario from (B), we can show that (c) needs to
* be paired with (d).
*
* D) exit_mm vs membarrier
*
* Two thread groups are created, A and B. Thread group B is created by
* issuing clone from group A with flag CLONE_VM set, but not CLONE_THREAD.
* Let's assume we have a single thread within each thread group (Thread A
* and Thread B). Thread A runs on CPU0, Thread B runs on CPU1.
*
* CPU0 CPU1
*
* membarrier():
* a: smp_mb()
* exit_mm():
* d: smp_mb()
* e: current->mm = NULL
* b: read rq->curr->mm == NULL
* c: smp_mb()
*
* Using scenario (B), we can show that (c) needs to be paired with (d).
*
* E) kthread_{use,unuse}_mm vs membarrier
*
* CPU0 CPU1
*
* membarrier():
* a: smp_mb()
* kthread_unuse_mm()
* d: smp_mb()
* e: current->mm = NULL
* b: read rq->curr->mm == NULL
* kthread_use_mm()
* f: current->mm = mm
* g: smp_mb()
* c: smp_mb()
*
* Using the scenario from (A), we can show that (a) needs to be paired
* with (g). Using the scenario from (B), we can show that (c) needs to
* be paired with (d).
*/
/* /*
* Bitmask made from a "or" of all commands within enum membarrier_cmd, * Bitmask made from a "or" of all commands within enum membarrier_cmd,
* except MEMBARRIER_CMD_QUERY. * except MEMBARRIER_CMD_QUERY.
......
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