[PATCH] kexec: change locking mechanism to a mutex

[Eric DeVolder]

On 9/21/23 19:26, Andrew Morton wrote:
> On Thu, 21 Sep 2023 17:59:38 -0400 Eric DeVolder <eric.devolder at oracle.com> wrote:
> 
>> Scaled up testing has revealed that the kexec_trylock()
>> implementation leads to failures within the crash hotplug
>> infrastructure due to the inability to acquire the lock,
>> specifically the message:
>>
>>   crash hp: kexec_trylock() failed, elfcorehdr may be inaccurate
>>
>> When hotplug events occur, the crash hotplug infrastructure first
>> attempts to obtain the lock via the kexec_trylock(). However, the
>> implementation either acquires the lock, or fails and returns; there
>> is no waiting on the lock. Here is the comment/explanation from
>> kernel/kexec_internal.h:kexec_trylock():
>>
>>   * Whatever is used to serialize accesses to the kexec_crash_image needs to be
>>   * NMI safe, as __crash_kexec() can happen during nmi_panic(), so here we use a
>>   * "simple" atomic variable that is acquired with a cmpxchg().
>>
>> While this in theory can happen for either CPU or memory hoptlug,
>> this problem is most prone to occur for memory hotplug.
>>
>> When memory is hot plugged, the memory is converted into smaller
>> 128MiB memblocks (typically). As each memblock is processed, a
>> kernel thread and a udev event thread are created. The udev thread
>> tries for the lock via the reading of the sysfs node
>> /sys/devices/system/memory/crash_hotplug node, and the kernel
>> worker thread tries for the lock upon entering the crash hotplug
>> infrastructure.
>>
>> These threads then compete for the kexec lock.
>>
>> For example, a 1GiB DIMM is converted into 8 memblocks, each
>> spawning two threads for a total of 16 threads that create a small
>> "swarm" all trying to acquire the lock. The larger the DIMM, the
>> more the memblocks and the larger the swarm.
>>
>> At the root of the problem is the atomic lock behind kexec_trylock();
>> it works well for low lock traffic; ie loading/unloading a capture
>> kernel, things that happen basically once. But with the introduction
>> of crash hotplug, the traffic through the lock increases significantly,
>> and more importantly in bursts occurring at roughly the same time. Thus
>> there is a need to wait on the lock.
>>
>> A possible workaround is to simply retry the lock, say up to N times.
>> There is, of course, the problem of determining a value of N that works for
>> all implementations, and for all the other call sites of kexec_trylock().
>> Not ideal.
>>
>> The design decision to use the atomic lock is described in the comment
>> from kexec_internal.h, cited above. However, examining the code of
>> __crash_kexec():
>>
>>          if (kexec_trylock()) {
>>                  if (kexec_crash_image) {
>>                          ...
>>                  }
>>                  kexec_unlock();
>>          }
>>
>> reveals that the use of kexec_trylock() here is actually a "best effort"
>> due to the atomic lock.  This atomic lock, prior to crash hotplug,
>> would almost always be assured (another kexec syscall could hold the lock
>> and prevent this, but that is about it).
>>
>> So at the point where the capture kernel would be invoked, if the lock
>> is not obtained, then kdump doesn't occur.
>>
>> It is possible to instead use a mutex with proper waiting, and utilize
>> mutex_trylock() as the "best effort" in __crash_kexec(). The use of a
>> mutex then avoids all the lock acquisition problems that were revealed
>> by the crash hotplug activity.
>>
>> Convert the atomic lock to a mutex.
>>
>> ...
>>
>> --- a/kernel/kexec_core.c
>> +++ b/kernel/kexec_core.c
>> @@ -47,7 +47,7 @@
>>   #include <crypto/hash.h>
>>   #include "kexec_internal.h"
>>   
>> -atomic_t __kexec_lock = ATOMIC_INIT(0);
>> +DEFINE_MUTEX(__kexec_lock);
>>   
>>   /* Flag to indicate we are going to kexec a new kernel */
>>   bool kexec_in_progress = false;
>> @@ -1057,7 +1057,7 @@ void __noclone __crash_kexec(struct pt_regs *regs)
>>   	 * of memory the xchg(&kexec_crash_image) would be
>>   	 * sufficient.  But since I reuse the memory...
>>   	 */
>> -	if (kexec_trylock()) {
>> +	if (mutex_trylock(&__kexec_lock)) {
>>   		if (kexec_crash_image) {
>>   			struct pt_regs fixed_regs;
> 
> What's happening here?  If someone else held the lock we silently fail
> to run the kexec?  Shouldn't we at least alert the user to what just
> happened?
> 
> 
Yes, I believe it would silently "fail" and not run the kexec kernel.
I do not have a good feel to know if logging is going to be functional,
and reliable, at this point in time (on a panic path)...
eric