linux/fs/btrfs/async-thread.c

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// SPDX-License-Identifier: GPL-2.0
/*
* Copyright (C) 2007 Oracle. All rights reserved.
* Copyright (C) 2014 Fujitsu. All rights reserved.
*/
#include <linux/kthread.h>
include cleanup: Update gfp.h and slab.h includes to prepare for breaking implicit slab.h inclusion from percpu.h percpu.h is included by sched.h and module.h and thus ends up being included when building most .c files. percpu.h includes slab.h which in turn includes gfp.h making everything defined by the two files universally available and complicating inclusion dependencies. percpu.h -> slab.h dependency is about to be removed. Prepare for this change by updating users of gfp and slab facilities include those headers directly instead of assuming availability. As this conversion needs to touch large number of source files, the following script is used as the basis of conversion. http://userweb.kernel.org/~tj/misc/slabh-sweep.py The script does the followings. * Scan files for gfp and slab usages and update includes such that only the necessary includes are there. ie. if only gfp is used, gfp.h, if slab is used, slab.h. * When the script inserts a new include, it looks at the include blocks and try to put the new include such that its order conforms to its surrounding. It's put in the include block which contains core kernel includes, in the same order that the rest are ordered - alphabetical, Christmas tree, rev-Xmas-tree or at the end if there doesn't seem to be any matching order. * If the script can't find a place to put a new include (mostly because the file doesn't have fitting include block), it prints out an error message indicating which .h file needs to be added to the file. The conversion was done in the following steps. 1. The initial automatic conversion of all .c files updated slightly over 4000 files, deleting around 700 includes and adding ~480 gfp.h and ~3000 slab.h inclusions. The script emitted errors for ~400 files. 2. Each error was manually checked. Some didn't need the inclusion, some needed manual addition while adding it to implementation .h or embedding .c file was more appropriate for others. This step added inclusions to around 150 files. 3. The script was run again and the output was compared to the edits from #2 to make sure no file was left behind. 4. Several build tests were done and a couple of problems were fixed. e.g. lib/decompress_*.c used malloc/free() wrappers around slab APIs requiring slab.h to be added manually. 5. The script was run on all .h files but without automatically editing them as sprinkling gfp.h and slab.h inclusions around .h files could easily lead to inclusion dependency hell. Most gfp.h inclusion directives were ignored as stuff from gfp.h was usually wildly available and often used in preprocessor macros. Each slab.h inclusion directive was examined and added manually as necessary. 6. percpu.h was updated not to include slab.h. 7. Build test were done on the following configurations and failures were fixed. CONFIG_GCOV_KERNEL was turned off for all tests (as my distributed build env didn't work with gcov compiles) and a few more options had to be turned off depending on archs to make things build (like ipr on powerpc/64 which failed due to missing writeq). * x86 and x86_64 UP and SMP allmodconfig and a custom test config. * powerpc and powerpc64 SMP allmodconfig * sparc and sparc64 SMP allmodconfig * ia64 SMP allmodconfig * s390 SMP allmodconfig * alpha SMP allmodconfig * um on x86_64 SMP allmodconfig 8. percpu.h modifications were reverted so that it could be applied as a separate patch and serve as bisection point. Given the fact that I had only a couple of failures from tests on step 6, I'm fairly confident about the coverage of this conversion patch. If there is a breakage, it's likely to be something in one of the arch headers which should be easily discoverable easily on most builds of the specific arch. Signed-off-by: Tejun Heo <tj@kernel.org> Guess-its-ok-by: Christoph Lameter <cl@linux-foundation.org> Cc: Ingo Molnar <mingo@redhat.com> Cc: Lee Schermerhorn <Lee.Schermerhorn@hp.com>
2010-03-24 08:04:11 +00:00
#include <linux/slab.h>
#include <linux/list.h>
#include <linux/spinlock.h>
#include <linux/freezer.h>
#include "async-thread.h"
#include "ctree.h"
enum {
WORK_DONE_BIT,
WORK_ORDER_DONE_BIT,
WORK_HIGH_PRIO_BIT,
};
Btrfs: Add ordered async work queues Btrfs uses kernel threads to create async work queues for cpu intensive operations such as checksumming and decompression. These work well, but they make it difficult to keep IO order intact. A single writepages call from pdflush or fsync will turn into a number of bios, and each bio is checksummed in parallel. Once the checksum is computed, the bio is sent down to the disk, and since we don't control the order in which the parallel operations happen, they might go down to the disk in almost any order. The code deals with this somewhat by having deep work queues for a single kernel thread, making it very likely that a single thread will process all the bios for a single inode. This patch introduces an explicitly ordered work queue. As work structs are placed into the queue they are put onto the tail of a list. They have three callbacks: ->func (cpu intensive processing here) ->ordered_func (order sensitive processing here) ->ordered_free (free the work struct, all processing is done) The work struct has three callbacks. The func callback does the cpu intensive work, and when it completes the work struct is marked as done. Every time a work struct completes, the list is checked to see if the head is marked as done. If so the ordered_func callback is used to do the order sensitive processing and the ordered_free callback is used to do any cleanup. Then we loop back and check the head of the list again. This patch also changes the checksumming code to use the ordered workqueues. One a 4 drive array, it increases streaming writes from 280MB/s to 350MB/s. Signed-off-by: Chris Mason <chris.mason@oracle.com>
2008-11-07 03:03:00 +00:00
#define NO_THRESHOLD (-1)
#define DFT_THRESHOLD (32)
struct __btrfs_workqueue {
struct workqueue_struct *normal_wq;
/* File system this workqueue services */
struct btrfs_fs_info *fs_info;
/* List head pointing to ordered work list */
struct list_head ordered_list;
/* Spinlock for ordered_list */
spinlock_t list_lock;
/* Thresholding related variants */
atomic_t pending;
/* Up limit of concurrency workers */
int limit_active;
/* Current number of concurrency workers */
int current_active;
/* Threshold to change current_active */
int thresh;
unsigned int count;
spinlock_t thres_lock;
};
struct btrfs_workqueue {
struct __btrfs_workqueue *normal;
struct __btrfs_workqueue *high;
};
struct btrfs_fs_info * __pure btrfs_workqueue_owner(const struct __btrfs_workqueue *wq)
{
return wq->fs_info;
}
struct btrfs_fs_info * __pure btrfs_work_owner(const struct btrfs_work *work)
{
return work->wq->fs_info;
}
bool btrfs_workqueue_normal_congested(const struct btrfs_workqueue *wq)
btrfs: limit async_work allocation and worker func duration Problem statement: unprivileged user who has read-write access to more than one btrfs subvolume may easily consume all kernel memory (eventually triggering oom-killer). Reproducer (./mkrmdir below essentially loops over mkdir/rmdir): [root@kteam1 ~]# cat prep.sh DEV=/dev/sdb mkfs.btrfs -f $DEV mount $DEV /mnt for i in `seq 1 16` do mkdir /mnt/$i btrfs subvolume create /mnt/SV_$i ID=`btrfs subvolume list /mnt |grep "SV_$i$" |cut -d ' ' -f 2` mount -t btrfs -o subvolid=$ID $DEV /mnt/$i chmod a+rwx /mnt/$i done [root@kteam1 ~]# sh prep.sh [maxim@kteam1 ~]$ for i in `seq 1 16`; do ./mkrmdir /mnt/$i 2000 2000 & done [root@kteam1 ~]# for i in `seq 1 4`; do grep "kmalloc-128" /proc/slabinfo | grep -v dma; sleep 60; done kmalloc-128 10144 10144 128 32 1 : tunables 0 0 0 : slabdata 317 317 0 kmalloc-128 9992352 9992352 128 32 1 : tunables 0 0 0 : slabdata 312261 312261 0 kmalloc-128 24226752 24226752 128 32 1 : tunables 0 0 0 : slabdata 757086 757086 0 kmalloc-128 42754240 42754240 128 32 1 : tunables 0 0 0 : slabdata 1336070 1336070 0 The huge numbers above come from insane number of async_work-s allocated and queued by btrfs_wq_run_delayed_node. The problem is caused by btrfs_wq_run_delayed_node() queuing more and more works if the number of delayed items is above BTRFS_DELAYED_BACKGROUND. The worker func (btrfs_async_run_delayed_root) processes at least BTRFS_DELAYED_BATCH items (if they are present in the list). So, the machinery works as expected while the list is almost empty. As soon as it is getting bigger, worker func starts to process more than one item at a time, it takes longer, and the chances to have async_works queued more than needed is getting higher. The problem above is worsened by another flaw of delayed-inode implementation: if async_work was queued in a throttling branch (number of items >= BTRFS_DELAYED_WRITEBACK), corresponding worker func won't quit until the number of items < BTRFS_DELAYED_BACKGROUND / 2. So, it is possible that the func occupies CPU infinitely (up to 30sec in my experiments): while the func is trying to drain the list, the user activity may add more and more items to the list. The patch fixes both problems in straightforward way: refuse queuing too many works in btrfs_wq_run_delayed_node and bail out of worker func if at least BTRFS_DELAYED_WRITEBACK items are processed. Changed in v2: remove support of thresh == NO_THRESHOLD. Signed-off-by: Maxim Patlasov <mpatlasov@virtuozzo.com> Signed-off-by: Chris Mason <clm@fb.com> Cc: stable@vger.kernel.org # v3.15+
2016-12-12 22:32:44 +00:00
{
/*
* We could compare wq->normal->pending with num_online_cpus()
* to support "thresh == NO_THRESHOLD" case, but it requires
* moving up atomic_inc/dec in thresh_queue/exec_hook. Let's
* postpone it until someone needs the support of that case.
*/
if (wq->normal->thresh == NO_THRESHOLD)
return false;
return atomic_read(&wq->normal->pending) > wq->normal->thresh * 2;
}
Btrfs: fix task hang under heavy compressed write This has been reported and discussed for a long time, and this hang occurs in both 3.15 and 3.16. Btrfs now migrates to use kernel workqueue, but it introduces this hang problem. Btrfs has a kind of work queued as an ordered way, which means that its ordered_func() must be processed in the way of FIFO, so it usually looks like -- normal_work_helper(arg) work = container_of(arg, struct btrfs_work, normal_work); work->func() <---- (we name it work X) for ordered_work in wq->ordered_list ordered_work->ordered_func() ordered_work->ordered_free() The hang is a rare case, first when we find free space, we get an uncached block group, then we go to read its free space cache inode for free space information, so it will file a readahead request btrfs_readpages() for page that is not in page cache __do_readpage() submit_extent_page() btrfs_submit_bio_hook() btrfs_bio_wq_end_io() submit_bio() end_workqueue_bio() <--(ret by the 1st endio) queue a work(named work Y) for the 2nd also the real endio() So the hang occurs when work Y's work_struct and work X's work_struct happens to share the same address. A bit more explanation, A,B,C -- struct btrfs_work arg -- struct work_struct kthread: worker_thread() pick up a work_struct from @worklist process_one_work(arg) worker->current_work = arg; <-- arg is A->normal_work worker->current_func(arg) normal_work_helper(arg) A = container_of(arg, struct btrfs_work, normal_work); A->func() A->ordered_func() A->ordered_free() <-- A gets freed B->ordered_func() submit_compressed_extents() find_free_extent() load_free_space_inode() ... <-- (the above readhead stack) end_workqueue_bio() btrfs_queue_work(work C) B->ordered_free() As if work A has a high priority in wq->ordered_list and there are more ordered works queued after it, such as B->ordered_func(), its memory could have been freed before normal_work_helper() returns, which means that kernel workqueue code worker_thread() still has worker->current_work pointer to be work A->normal_work's, ie. arg's address. Meanwhile, work C is allocated after work A is freed, work C->normal_work and work A->normal_work are likely to share the same address(I confirmed this with ftrace output, so I'm not just guessing, it's rare though). When another kthread picks up work C->normal_work to process, and finds our kthread is processing it(see find_worker_executing_work()), it'll think work C as a collision and skip then, which ends up nobody processing work C. So the situation is that our kthread is waiting forever on work C. Besides, there're other cases that can lead to deadlock, but the real problem is that all btrfs workqueue shares one work->func, -- normal_work_helper, so this makes each workqueue to have its own helper function, but only a wraper pf normal_work_helper. With this patch, I no long hit the above hang. Signed-off-by: Liu Bo <bo.li.liu@oracle.com> Signed-off-by: Chris Mason <clm@fb.com>
2014-08-15 15:36:53 +00:00
static struct __btrfs_workqueue *
__btrfs_alloc_workqueue(struct btrfs_fs_info *fs_info, const char *name,
unsigned int flags, int limit_active, int thresh)
{
struct __btrfs_workqueue *ret = kzalloc(sizeof(*ret), GFP_KERNEL);
if (!ret)
return NULL;
ret->fs_info = fs_info;
ret->limit_active = limit_active;
atomic_set(&ret->pending, 0);
if (thresh == 0)
thresh = DFT_THRESHOLD;
/* For low threshold, disabling threshold is a better choice */
if (thresh < DFT_THRESHOLD) {
ret->current_active = limit_active;
ret->thresh = NO_THRESHOLD;
} else {
/*
* For threshold-able wq, let its concurrency grow on demand.
* Use minimal max_active at alloc time to reduce resource
* usage.
*/
ret->current_active = 1;
ret->thresh = thresh;
}
if (flags & WQ_HIGHPRI)
ret->normal_wq = alloc_workqueue("btrfs-%s-high", flags,
ret->current_active, name);
else
ret->normal_wq = alloc_workqueue("btrfs-%s", flags,
ret->current_active, name);
if (!ret->normal_wq) {
kfree(ret);
return NULL;
}
INIT_LIST_HEAD(&ret->ordered_list);
spin_lock_init(&ret->list_lock);
spin_lock_init(&ret->thres_lock);
trace_btrfs_workqueue_alloc(ret, name, flags & WQ_HIGHPRI);
return ret;
}
static inline void
__btrfs_destroy_workqueue(struct __btrfs_workqueue *wq);
struct btrfs_workqueue *btrfs_alloc_workqueue(struct btrfs_fs_info *fs_info,
const char *name,
unsigned int flags,
int limit_active,
int thresh)
{
struct btrfs_workqueue *ret = kzalloc(sizeof(*ret), GFP_KERNEL);
if (!ret)
return NULL;
ret->normal = __btrfs_alloc_workqueue(fs_info, name,
flags & ~WQ_HIGHPRI,
limit_active, thresh);
if (!ret->normal) {
kfree(ret);
return NULL;
}
if (flags & WQ_HIGHPRI) {
ret->high = __btrfs_alloc_workqueue(fs_info, name, flags,
limit_active, thresh);
if (!ret->high) {
__btrfs_destroy_workqueue(ret->normal);
kfree(ret);
return NULL;
}
}
return ret;
}
/*
* Hook for threshold which will be called in btrfs_queue_work.
* This hook WILL be called in IRQ handler context,
* so workqueue_set_max_active MUST NOT be called in this hook
*/
static inline void thresh_queue_hook(struct __btrfs_workqueue *wq)
{
if (wq->thresh == NO_THRESHOLD)
return;
atomic_inc(&wq->pending);
}
/*
* Hook for threshold which will be called before executing the work,
* This hook is called in kthread content.
* So workqueue_set_max_active is called here.
*/
static inline void thresh_exec_hook(struct __btrfs_workqueue *wq)
{
int new_current_active;
long pending;
int need_change = 0;
if (wq->thresh == NO_THRESHOLD)
return;
atomic_dec(&wq->pending);
spin_lock(&wq->thres_lock);
/*
* Use wq->count to limit the calling frequency of
* workqueue_set_max_active.
*/
wq->count++;
wq->count %= (wq->thresh / 4);
if (!wq->count)
goto out;
new_current_active = wq->current_active;
/*
* pending may be changed later, but it's OK since we really
* don't need it so accurate to calculate new_max_active.
*/
pending = atomic_read(&wq->pending);
if (pending > wq->thresh)
new_current_active++;
if (pending < wq->thresh / 2)
new_current_active--;
new_current_active = clamp_val(new_current_active, 1, wq->limit_active);
if (new_current_active != wq->current_active) {
need_change = 1;
wq->current_active = new_current_active;
}
out:
spin_unlock(&wq->thres_lock);
if (need_change) {
workqueue_set_max_active(wq->normal_wq, wq->current_active);
}
}
btrfs: don't prematurely free work in run_ordered_work() We hit the following very strange deadlock on a system with Btrfs on a loop device backed by another Btrfs filesystem: 1. The top (loop device) filesystem queues an async_cow work item from cow_file_range_async(). We'll call this work X. 2. Worker thread A starts work X (normal_work_helper()). 3. Worker thread A executes the ordered work for the top filesystem (run_ordered_work()). 4. Worker thread A finishes the ordered work for work X and frees X (work->ordered_free()). 5. Worker thread A executes another ordered work and gets blocked on I/O to the bottom filesystem (still in run_ordered_work()). 6. Meanwhile, the bottom filesystem allocates and queues an async_cow work item which happens to be the recently-freed X. 7. The workqueue code sees that X is already being executed by worker thread A, so it schedules X to be executed _after_ worker thread A finishes (see the find_worker_executing_work() call in process_one_work()). Now, the top filesystem is waiting for I/O on the bottom filesystem, but the bottom filesystem is waiting for the top filesystem to finish, so we deadlock. This happens because we are breaking the workqueue assumption that a work item cannot be recycled while it still depends on other work. Fix it by waiting to free the work item until we are done with all of the related ordered work. P.S.: One might ask why the workqueue code doesn't try to detect a recycled work item. It actually does try by checking whether the work item has the same work function (find_worker_executing_work()), but in our case the function is the same. This is the only key that the workqueue code has available to compare, short of adding an additional, layer-violating "custom key". Considering that we're the only ones that have ever hit this, we should just play by the rules. Unfortunately, we haven't been able to create a minimal reproducer other than our full container setup using a compress-force=zstd filesystem on top of another compress-force=zstd filesystem. Suggested-by: Tejun Heo <tj@kernel.org> Reviewed-by: Johannes Thumshirn <jthumshirn@suse.de> Signed-off-by: Omar Sandoval <osandov@fb.com> Reviewed-by: David Sterba <dsterba@suse.com> Signed-off-by: David Sterba <dsterba@suse.com>
2019-09-16 18:30:53 +00:00
static void run_ordered_work(struct __btrfs_workqueue *wq,
struct btrfs_work *self)
{
struct list_head *list = &wq->ordered_list;
struct btrfs_work *work;
spinlock_t *lock = &wq->list_lock;
unsigned long flags;
btrfs: don't prematurely free work in run_ordered_work() We hit the following very strange deadlock on a system with Btrfs on a loop device backed by another Btrfs filesystem: 1. The top (loop device) filesystem queues an async_cow work item from cow_file_range_async(). We'll call this work X. 2. Worker thread A starts work X (normal_work_helper()). 3. Worker thread A executes the ordered work for the top filesystem (run_ordered_work()). 4. Worker thread A finishes the ordered work for work X and frees X (work->ordered_free()). 5. Worker thread A executes another ordered work and gets blocked on I/O to the bottom filesystem (still in run_ordered_work()). 6. Meanwhile, the bottom filesystem allocates and queues an async_cow work item which happens to be the recently-freed X. 7. The workqueue code sees that X is already being executed by worker thread A, so it schedules X to be executed _after_ worker thread A finishes (see the find_worker_executing_work() call in process_one_work()). Now, the top filesystem is waiting for I/O on the bottom filesystem, but the bottom filesystem is waiting for the top filesystem to finish, so we deadlock. This happens because we are breaking the workqueue assumption that a work item cannot be recycled while it still depends on other work. Fix it by waiting to free the work item until we are done with all of the related ordered work. P.S.: One might ask why the workqueue code doesn't try to detect a recycled work item. It actually does try by checking whether the work item has the same work function (find_worker_executing_work()), but in our case the function is the same. This is the only key that the workqueue code has available to compare, short of adding an additional, layer-violating "custom key". Considering that we're the only ones that have ever hit this, we should just play by the rules. Unfortunately, we haven't been able to create a minimal reproducer other than our full container setup using a compress-force=zstd filesystem on top of another compress-force=zstd filesystem. Suggested-by: Tejun Heo <tj@kernel.org> Reviewed-by: Johannes Thumshirn <jthumshirn@suse.de> Signed-off-by: Omar Sandoval <osandov@fb.com> Reviewed-by: David Sterba <dsterba@suse.com> Signed-off-by: David Sterba <dsterba@suse.com>
2019-09-16 18:30:53 +00:00
bool free_self = false;
while (1) {
spin_lock_irqsave(lock, flags);
if (list_empty(list))
break;
work = list_entry(list->next, struct btrfs_work,
ordered_list);
if (!test_bit(WORK_DONE_BIT, &work->flags))
break;
/*
* we are going to call the ordered done function, but
* we leave the work item on the list as a barrier so
* that later work items that are done don't have their
* functions called before this one returns
*/
if (test_and_set_bit(WORK_ORDER_DONE_BIT, &work->flags))
break;
trace_btrfs_ordered_sched(work);
spin_unlock_irqrestore(lock, flags);
work->ordered_func(work);
/* now take the lock again and drop our item from the list */
spin_lock_irqsave(lock, flags);
list_del(&work->ordered_list);
spin_unlock_irqrestore(lock, flags);
btrfs: don't prematurely free work in run_ordered_work() We hit the following very strange deadlock on a system with Btrfs on a loop device backed by another Btrfs filesystem: 1. The top (loop device) filesystem queues an async_cow work item from cow_file_range_async(). We'll call this work X. 2. Worker thread A starts work X (normal_work_helper()). 3. Worker thread A executes the ordered work for the top filesystem (run_ordered_work()). 4. Worker thread A finishes the ordered work for work X and frees X (work->ordered_free()). 5. Worker thread A executes another ordered work and gets blocked on I/O to the bottom filesystem (still in run_ordered_work()). 6. Meanwhile, the bottom filesystem allocates and queues an async_cow work item which happens to be the recently-freed X. 7. The workqueue code sees that X is already being executed by worker thread A, so it schedules X to be executed _after_ worker thread A finishes (see the find_worker_executing_work() call in process_one_work()). Now, the top filesystem is waiting for I/O on the bottom filesystem, but the bottom filesystem is waiting for the top filesystem to finish, so we deadlock. This happens because we are breaking the workqueue assumption that a work item cannot be recycled while it still depends on other work. Fix it by waiting to free the work item until we are done with all of the related ordered work. P.S.: One might ask why the workqueue code doesn't try to detect a recycled work item. It actually does try by checking whether the work item has the same work function (find_worker_executing_work()), but in our case the function is the same. This is the only key that the workqueue code has available to compare, short of adding an additional, layer-violating "custom key". Considering that we're the only ones that have ever hit this, we should just play by the rules. Unfortunately, we haven't been able to create a minimal reproducer other than our full container setup using a compress-force=zstd filesystem on top of another compress-force=zstd filesystem. Suggested-by: Tejun Heo <tj@kernel.org> Reviewed-by: Johannes Thumshirn <jthumshirn@suse.de> Signed-off-by: Omar Sandoval <osandov@fb.com> Reviewed-by: David Sterba <dsterba@suse.com> Signed-off-by: David Sterba <dsterba@suse.com>
2019-09-16 18:30:53 +00:00
if (work == self) {
/*
* This is the work item that the worker is currently
* executing.
*
* The kernel workqueue code guarantees non-reentrancy
* of work items. I.e., if a work item with the same
* address and work function is queued twice, the second
* execution is blocked until the first one finishes. A
* work item may be freed and recycled with the same
* work function; the workqueue code assumes that the
* original work item cannot depend on the recycled work
* item in that case (see find_worker_executing_work()).
*
* Note that different types of Btrfs work can depend on
* each other, and one type of work on one Btrfs
* filesystem may even depend on the same type of work
* on another Btrfs filesystem via, e.g., a loop device.
* Therefore, we must not allow the current work item to
* be recycled until we are really done, otherwise we
* break the above assumption and can deadlock.
btrfs: don't prematurely free work in run_ordered_work() We hit the following very strange deadlock on a system with Btrfs on a loop device backed by another Btrfs filesystem: 1. The top (loop device) filesystem queues an async_cow work item from cow_file_range_async(). We'll call this work X. 2. Worker thread A starts work X (normal_work_helper()). 3. Worker thread A executes the ordered work for the top filesystem (run_ordered_work()). 4. Worker thread A finishes the ordered work for work X and frees X (work->ordered_free()). 5. Worker thread A executes another ordered work and gets blocked on I/O to the bottom filesystem (still in run_ordered_work()). 6. Meanwhile, the bottom filesystem allocates and queues an async_cow work item which happens to be the recently-freed X. 7. The workqueue code sees that X is already being executed by worker thread A, so it schedules X to be executed _after_ worker thread A finishes (see the find_worker_executing_work() call in process_one_work()). Now, the top filesystem is waiting for I/O on the bottom filesystem, but the bottom filesystem is waiting for the top filesystem to finish, so we deadlock. This happens because we are breaking the workqueue assumption that a work item cannot be recycled while it still depends on other work. Fix it by waiting to free the work item until we are done with all of the related ordered work. P.S.: One might ask why the workqueue code doesn't try to detect a recycled work item. It actually does try by checking whether the work item has the same work function (find_worker_executing_work()), but in our case the function is the same. This is the only key that the workqueue code has available to compare, short of adding an additional, layer-violating "custom key". Considering that we're the only ones that have ever hit this, we should just play by the rules. Unfortunately, we haven't been able to create a minimal reproducer other than our full container setup using a compress-force=zstd filesystem on top of another compress-force=zstd filesystem. Suggested-by: Tejun Heo <tj@kernel.org> Reviewed-by: Johannes Thumshirn <jthumshirn@suse.de> Signed-off-by: Omar Sandoval <osandov@fb.com> Reviewed-by: David Sterba <dsterba@suse.com> Signed-off-by: David Sterba <dsterba@suse.com>
2019-09-16 18:30:53 +00:00
*/
free_self = true;
} else {
/*
* We don't want to call the ordered free functions with
* the lock held.
btrfs: don't prematurely free work in run_ordered_work() We hit the following very strange deadlock on a system with Btrfs on a loop device backed by another Btrfs filesystem: 1. The top (loop device) filesystem queues an async_cow work item from cow_file_range_async(). We'll call this work X. 2. Worker thread A starts work X (normal_work_helper()). 3. Worker thread A executes the ordered work for the top filesystem (run_ordered_work()). 4. Worker thread A finishes the ordered work for work X and frees X (work->ordered_free()). 5. Worker thread A executes another ordered work and gets blocked on I/O to the bottom filesystem (still in run_ordered_work()). 6. Meanwhile, the bottom filesystem allocates and queues an async_cow work item which happens to be the recently-freed X. 7. The workqueue code sees that X is already being executed by worker thread A, so it schedules X to be executed _after_ worker thread A finishes (see the find_worker_executing_work() call in process_one_work()). Now, the top filesystem is waiting for I/O on the bottom filesystem, but the bottom filesystem is waiting for the top filesystem to finish, so we deadlock. This happens because we are breaking the workqueue assumption that a work item cannot be recycled while it still depends on other work. Fix it by waiting to free the work item until we are done with all of the related ordered work. P.S.: One might ask why the workqueue code doesn't try to detect a recycled work item. It actually does try by checking whether the work item has the same work function (find_worker_executing_work()), but in our case the function is the same. This is the only key that the workqueue code has available to compare, short of adding an additional, layer-violating "custom key". Considering that we're the only ones that have ever hit this, we should just play by the rules. Unfortunately, we haven't been able to create a minimal reproducer other than our full container setup using a compress-force=zstd filesystem on top of another compress-force=zstd filesystem. Suggested-by: Tejun Heo <tj@kernel.org> Reviewed-by: Johannes Thumshirn <jthumshirn@suse.de> Signed-off-by: Omar Sandoval <osandov@fb.com> Reviewed-by: David Sterba <dsterba@suse.com> Signed-off-by: David Sterba <dsterba@suse.com>
2019-09-16 18:30:53 +00:00
*/
work->ordered_free(work);
/* NB: work must not be dereferenced past this point. */
trace_btrfs_all_work_done(wq->fs_info, work);
btrfs: don't prematurely free work in run_ordered_work() We hit the following very strange deadlock on a system with Btrfs on a loop device backed by another Btrfs filesystem: 1. The top (loop device) filesystem queues an async_cow work item from cow_file_range_async(). We'll call this work X. 2. Worker thread A starts work X (normal_work_helper()). 3. Worker thread A executes the ordered work for the top filesystem (run_ordered_work()). 4. Worker thread A finishes the ordered work for work X and frees X (work->ordered_free()). 5. Worker thread A executes another ordered work and gets blocked on I/O to the bottom filesystem (still in run_ordered_work()). 6. Meanwhile, the bottom filesystem allocates and queues an async_cow work item which happens to be the recently-freed X. 7. The workqueue code sees that X is already being executed by worker thread A, so it schedules X to be executed _after_ worker thread A finishes (see the find_worker_executing_work() call in process_one_work()). Now, the top filesystem is waiting for I/O on the bottom filesystem, but the bottom filesystem is waiting for the top filesystem to finish, so we deadlock. This happens because we are breaking the workqueue assumption that a work item cannot be recycled while it still depends on other work. Fix it by waiting to free the work item until we are done with all of the related ordered work. P.S.: One might ask why the workqueue code doesn't try to detect a recycled work item. It actually does try by checking whether the work item has the same work function (find_worker_executing_work()), but in our case the function is the same. This is the only key that the workqueue code has available to compare, short of adding an additional, layer-violating "custom key". Considering that we're the only ones that have ever hit this, we should just play by the rules. Unfortunately, we haven't been able to create a minimal reproducer other than our full container setup using a compress-force=zstd filesystem on top of another compress-force=zstd filesystem. Suggested-by: Tejun Heo <tj@kernel.org> Reviewed-by: Johannes Thumshirn <jthumshirn@suse.de> Signed-off-by: Omar Sandoval <osandov@fb.com> Reviewed-by: David Sterba <dsterba@suse.com> Signed-off-by: David Sterba <dsterba@suse.com>
2019-09-16 18:30:53 +00:00
}
}
spin_unlock_irqrestore(lock, flags);
btrfs: don't prematurely free work in run_ordered_work() We hit the following very strange deadlock on a system with Btrfs on a loop device backed by another Btrfs filesystem: 1. The top (loop device) filesystem queues an async_cow work item from cow_file_range_async(). We'll call this work X. 2. Worker thread A starts work X (normal_work_helper()). 3. Worker thread A executes the ordered work for the top filesystem (run_ordered_work()). 4. Worker thread A finishes the ordered work for work X and frees X (work->ordered_free()). 5. Worker thread A executes another ordered work and gets blocked on I/O to the bottom filesystem (still in run_ordered_work()). 6. Meanwhile, the bottom filesystem allocates and queues an async_cow work item which happens to be the recently-freed X. 7. The workqueue code sees that X is already being executed by worker thread A, so it schedules X to be executed _after_ worker thread A finishes (see the find_worker_executing_work() call in process_one_work()). Now, the top filesystem is waiting for I/O on the bottom filesystem, but the bottom filesystem is waiting for the top filesystem to finish, so we deadlock. This happens because we are breaking the workqueue assumption that a work item cannot be recycled while it still depends on other work. Fix it by waiting to free the work item until we are done with all of the related ordered work. P.S.: One might ask why the workqueue code doesn't try to detect a recycled work item. It actually does try by checking whether the work item has the same work function (find_worker_executing_work()), but in our case the function is the same. This is the only key that the workqueue code has available to compare, short of adding an additional, layer-violating "custom key". Considering that we're the only ones that have ever hit this, we should just play by the rules. Unfortunately, we haven't been able to create a minimal reproducer other than our full container setup using a compress-force=zstd filesystem on top of another compress-force=zstd filesystem. Suggested-by: Tejun Heo <tj@kernel.org> Reviewed-by: Johannes Thumshirn <jthumshirn@suse.de> Signed-off-by: Omar Sandoval <osandov@fb.com> Reviewed-by: David Sterba <dsterba@suse.com> Signed-off-by: David Sterba <dsterba@suse.com>
2019-09-16 18:30:53 +00:00
if (free_self) {
self->ordered_free(self);
/* NB: self must not be dereferenced past this point. */
trace_btrfs_all_work_done(wq->fs_info, self);
btrfs: don't prematurely free work in run_ordered_work() We hit the following very strange deadlock on a system with Btrfs on a loop device backed by another Btrfs filesystem: 1. The top (loop device) filesystem queues an async_cow work item from cow_file_range_async(). We'll call this work X. 2. Worker thread A starts work X (normal_work_helper()). 3. Worker thread A executes the ordered work for the top filesystem (run_ordered_work()). 4. Worker thread A finishes the ordered work for work X and frees X (work->ordered_free()). 5. Worker thread A executes another ordered work and gets blocked on I/O to the bottom filesystem (still in run_ordered_work()). 6. Meanwhile, the bottom filesystem allocates and queues an async_cow work item which happens to be the recently-freed X. 7. The workqueue code sees that X is already being executed by worker thread A, so it schedules X to be executed _after_ worker thread A finishes (see the find_worker_executing_work() call in process_one_work()). Now, the top filesystem is waiting for I/O on the bottom filesystem, but the bottom filesystem is waiting for the top filesystem to finish, so we deadlock. This happens because we are breaking the workqueue assumption that a work item cannot be recycled while it still depends on other work. Fix it by waiting to free the work item until we are done with all of the related ordered work. P.S.: One might ask why the workqueue code doesn't try to detect a recycled work item. It actually does try by checking whether the work item has the same work function (find_worker_executing_work()), but in our case the function is the same. This is the only key that the workqueue code has available to compare, short of adding an additional, layer-violating "custom key". Considering that we're the only ones that have ever hit this, we should just play by the rules. Unfortunately, we haven't been able to create a minimal reproducer other than our full container setup using a compress-force=zstd filesystem on top of another compress-force=zstd filesystem. Suggested-by: Tejun Heo <tj@kernel.org> Reviewed-by: Johannes Thumshirn <jthumshirn@suse.de> Signed-off-by: Omar Sandoval <osandov@fb.com> Reviewed-by: David Sterba <dsterba@suse.com> Signed-off-by: David Sterba <dsterba@suse.com>
2019-09-16 18:30:53 +00:00
}
}
static void btrfs_work_helper(struct work_struct *normal_work)
{
struct btrfs_work *work = container_of(normal_work, struct btrfs_work,
normal_work);
struct __btrfs_workqueue *wq;
int need_order = 0;
/*
* We should not touch things inside work in the following cases:
* 1) after work->func() if it has no ordered_free
* Since the struct is freed in work->func().
* 2) after setting WORK_DONE_BIT
* The work may be freed in other threads almost instantly.
* So we save the needed things here.
*/
if (work->ordered_func)
need_order = 1;
wq = work->wq;
trace_btrfs_work_sched(work);
thresh_exec_hook(wq);
work->func(work);
if (need_order) {
set_bit(WORK_DONE_BIT, &work->flags);
btrfs: don't prematurely free work in run_ordered_work() We hit the following very strange deadlock on a system with Btrfs on a loop device backed by another Btrfs filesystem: 1. The top (loop device) filesystem queues an async_cow work item from cow_file_range_async(). We'll call this work X. 2. Worker thread A starts work X (normal_work_helper()). 3. Worker thread A executes the ordered work for the top filesystem (run_ordered_work()). 4. Worker thread A finishes the ordered work for work X and frees X (work->ordered_free()). 5. Worker thread A executes another ordered work and gets blocked on I/O to the bottom filesystem (still in run_ordered_work()). 6. Meanwhile, the bottom filesystem allocates and queues an async_cow work item which happens to be the recently-freed X. 7. The workqueue code sees that X is already being executed by worker thread A, so it schedules X to be executed _after_ worker thread A finishes (see the find_worker_executing_work() call in process_one_work()). Now, the top filesystem is waiting for I/O on the bottom filesystem, but the bottom filesystem is waiting for the top filesystem to finish, so we deadlock. This happens because we are breaking the workqueue assumption that a work item cannot be recycled while it still depends on other work. Fix it by waiting to free the work item until we are done with all of the related ordered work. P.S.: One might ask why the workqueue code doesn't try to detect a recycled work item. It actually does try by checking whether the work item has the same work function (find_worker_executing_work()), but in our case the function is the same. This is the only key that the workqueue code has available to compare, short of adding an additional, layer-violating "custom key". Considering that we're the only ones that have ever hit this, we should just play by the rules. Unfortunately, we haven't been able to create a minimal reproducer other than our full container setup using a compress-force=zstd filesystem on top of another compress-force=zstd filesystem. Suggested-by: Tejun Heo <tj@kernel.org> Reviewed-by: Johannes Thumshirn <jthumshirn@suse.de> Signed-off-by: Omar Sandoval <osandov@fb.com> Reviewed-by: David Sterba <dsterba@suse.com> Signed-off-by: David Sterba <dsterba@suse.com>
2019-09-16 18:30:53 +00:00
run_ordered_work(wq, work);
} else {
/* NB: work must not be dereferenced past this point. */
trace_btrfs_all_work_done(wq->fs_info, work);
}
}
void btrfs_init_work(struct btrfs_work *work, btrfs_func_t func,
btrfs_func_t ordered_func, btrfs_func_t ordered_free)
{
work->func = func;
work->ordered_func = ordered_func;
work->ordered_free = ordered_free;
INIT_WORK(&work->normal_work, btrfs_work_helper);
INIT_LIST_HEAD(&work->ordered_list);
work->flags = 0;
}
static inline void __btrfs_queue_work(struct __btrfs_workqueue *wq,
struct btrfs_work *work)
{
unsigned long flags;
work->wq = wq;
thresh_queue_hook(wq);
if (work->ordered_func) {
spin_lock_irqsave(&wq->list_lock, flags);
list_add_tail(&work->ordered_list, &wq->ordered_list);
spin_unlock_irqrestore(&wq->list_lock, flags);
}
trace_btrfs_work_queued(work);
queue_work(wq->normal_wq, &work->normal_work);
}
void btrfs_queue_work(struct btrfs_workqueue *wq,
struct btrfs_work *work)
{
struct __btrfs_workqueue *dest_wq;
if (test_bit(WORK_HIGH_PRIO_BIT, &work->flags) && wq->high)
dest_wq = wq->high;
else
dest_wq = wq->normal;
__btrfs_queue_work(dest_wq, work);
}
static inline void
__btrfs_destroy_workqueue(struct __btrfs_workqueue *wq)
{
destroy_workqueue(wq->normal_wq);
trace_btrfs_workqueue_destroy(wq);
kfree(wq);
}
void btrfs_destroy_workqueue(struct btrfs_workqueue *wq)
{
if (!wq)
return;
if (wq->high)
__btrfs_destroy_workqueue(wq->high);
__btrfs_destroy_workqueue(wq->normal);
kfree(wq);
}
void btrfs_workqueue_set_max(struct btrfs_workqueue *wq, int limit_active)
{
if (!wq)
return;
wq->normal->limit_active = limit_active;
if (wq->high)
wq->high->limit_active = limit_active;
}
void btrfs_set_work_high_priority(struct btrfs_work *work)
{
set_bit(WORK_HIGH_PRIO_BIT, &work->flags);
}
Btrfs: fix crash during unmount due to race with delayed inode workers During unmount we can have a job from the delayed inode items work queue still running, that can lead to at least two bad things: 1) A crash, because the worker can try to create a transaction just after the fs roots were freed; 2) A transaction leak, because the worker can create a transaction before the fs roots are freed and just after we committed the last transaction and after we stopped the transaction kthread. A stack trace example of the crash: [79011.691214] kernel BUG at lib/radix-tree.c:982! [79011.692056] invalid opcode: 0000 [#1] PREEMPT SMP DEBUG_PAGEALLOC PTI [79011.693180] CPU: 3 PID: 1394 Comm: kworker/u8:2 Tainted: G W 5.6.0-rc2-btrfs-next-54 #2 (...) [79011.696789] Workqueue: btrfs-delayed-meta btrfs_work_helper [btrfs] [79011.697904] RIP: 0010:radix_tree_tag_set+0xe7/0x170 (...) [79011.702014] RSP: 0018:ffffb3c84a317ca0 EFLAGS: 00010293 [79011.702949] RAX: 0000000000000000 RBX: 0000000000000000 RCX: 0000000000000000 [79011.704202] RDX: ffffb3c84a317cb0 RSI: ffffb3c84a317ca8 RDI: ffff8db3931340a0 [79011.705463] RBP: 0000000000000005 R08: 0000000000000005 R09: ffffffff974629d0 [79011.706756] R10: ffffb3c84a317bc0 R11: 0000000000000001 R12: ffff8db393134000 [79011.708010] R13: ffff8db3931340a0 R14: ffff8db393134068 R15: 0000000000000001 [79011.709270] FS: 0000000000000000(0000) GS:ffff8db3b6a00000(0000) knlGS:0000000000000000 [79011.710699] CS: 0010 DS: 0000 ES: 0000 CR0: 0000000080050033 [79011.711710] CR2: 00007f22c2a0a000 CR3: 0000000232ad4005 CR4: 00000000003606e0 [79011.712958] DR0: 0000000000000000 DR1: 0000000000000000 DR2: 0000000000000000 [79011.714205] DR3: 0000000000000000 DR6: 00000000fffe0ff0 DR7: 0000000000000400 [79011.715448] Call Trace: [79011.715925] record_root_in_trans+0x72/0xf0 [btrfs] [79011.716819] btrfs_record_root_in_trans+0x4b/0x70 [btrfs] [79011.717925] start_transaction+0xdd/0x5c0 [btrfs] [79011.718829] btrfs_async_run_delayed_root+0x17e/0x2b0 [btrfs] [79011.719915] btrfs_work_helper+0xaa/0x720 [btrfs] [79011.720773] process_one_work+0x26d/0x6a0 [79011.721497] worker_thread+0x4f/0x3e0 [79011.722153] ? process_one_work+0x6a0/0x6a0 [79011.722901] kthread+0x103/0x140 [79011.723481] ? kthread_create_worker_on_cpu+0x70/0x70 [79011.724379] ret_from_fork+0x3a/0x50 (...) The following diagram shows a sequence of steps that lead to the crash during ummount of the filesystem: CPU 1 CPU 2 CPU 3 btrfs_punch_hole() btrfs_btree_balance_dirty() btrfs_balance_delayed_items() --> sees fs_info->delayed_root->items with value 200, which is greater than BTRFS_DELAYED_BACKGROUND (128) and smaller than BTRFS_DELAYED_WRITEBACK (512) btrfs_wq_run_delayed_node() --> queues a job for fs_info->delayed_workers to run btrfs_async_run_delayed_root() btrfs_async_run_delayed_root() --> job queued by CPU 1 --> starts picking and running delayed nodes from the prepare_list list close_ctree() btrfs_delete_unused_bgs() btrfs_commit_super() btrfs_join_transaction() --> gets transaction N btrfs_commit_transaction(N) --> set transaction state to TRANTS_STATE_COMMIT_START btrfs_first_prepared_delayed_node() --> picks delayed node X through the prepared_list list btrfs_run_delayed_items() btrfs_first_delayed_node() --> also picks delayed node X but through the node_list list __btrfs_commit_inode_delayed_items() --> runs all delayed items from this node and drops the node's item count to 0 through call to btrfs_release_delayed_inode() --> finishes running any remaining delayed nodes --> finishes transaction commit --> stops cleaner and transaction threads btrfs_free_fs_roots() --> frees all roots and removes them from the radix tree fs_info->fs_roots_radix btrfs_join_transaction() start_transaction() btrfs_record_root_in_trans() record_root_in_trans() radix_tree_tag_set() --> crashes because the root is not in the radix tree anymore If the worker is able to call btrfs_join_transaction() before the unmount task frees the fs roots, we end up leaking a transaction and all its resources, since after the call to btrfs_commit_super() and stopping the transaction kthread, we don't expect to have any transaction open anymore. When this situation happens the worker has a delayed node that has no more items to run, since the task calling btrfs_run_delayed_items(), which is doing a transaction commit, picks the same node and runs all its items first. We can not wait for the worker to complete when running delayed items through btrfs_run_delayed_items(), because we call that function in several phases of a transaction commit, and that could cause a deadlock because the worker calls btrfs_join_transaction() and the task doing the transaction commit may have already set the transaction state to TRANS_STATE_COMMIT_DOING. Also it's not possible to get into a situation where only some of the items of a delayed node are added to the fs/subvolume tree in the current transaction and the remaining ones in the next transaction, because when running the items of a delayed inode we lock its mutex, effectively waiting for the worker if the worker is running the items of the delayed node already. Since this can only cause issues when unmounting a filesystem, fix it in a simple way by waiting for any jobs on the delayed workers queue before calling btrfs_commit_supper() at close_ctree(). This works because at this point no one can call btrfs_btree_balance_dirty() or btrfs_balance_delayed_items(), and if we end up waiting for any worker to complete, btrfs_commit_super() will commit the transaction created by the worker. CC: stable@vger.kernel.org # 4.4+ Signed-off-by: Filipe Manana <fdmanana@suse.com> Reviewed-by: David Sterba <dsterba@suse.com> Signed-off-by: David Sterba <dsterba@suse.com>
2020-02-28 13:04:36 +00:00
void btrfs_flush_workqueue(struct btrfs_workqueue *wq)
{
if (wq->high)
flush_workqueue(wq->high->normal_wq);
flush_workqueue(wq->normal->normal_wq);
}