Commit dbb26055 authored Nov 30, 2016 by Thomas Gleixner Committed by Ingo Molnar Dec 02, 2016

locking/rtmutex: Prevent dequeue vs. unlock race



David reported a futex/rtmutex state corruption. It's caused by the
following problem:

CPU0		CPU1		CPU2

l->owner=T1
		rt_mutex_lock(l)
		lock(l->wait_lock)
		l->owner = T1 | HAS_WAITERS;
		enqueue(T2)
		boost()
		  unlock(l->wait_lock)
		schedule()

				rt_mutex_lock(l)
				lock(l->wait_lock)
				l->owner = T1 | HAS_WAITERS;
				enqueue(T3)
				boost()
				  unlock(l->wait_lock)
				schedule()
		signal(->T2)	signal(->T3)
		lock(l->wait_lock)
		dequeue(T2)
		deboost()
		  unlock(l->wait_lock)
				lock(l->wait_lock)
				dequeue(T3)
				  ===> wait list is now empty
				deboost()
				 unlock(l->wait_lock)
		lock(l->wait_lock)
		fixup_rt_mutex_waiters()
		  if (wait_list_empty(l)) {
		    owner = l->owner & ~HAS_WAITERS;
		    l->owner = owner
		     ==> l->owner = T1
		  }

				lock(l->wait_lock)
rt_mutex_unlock(l)		fixup_rt_mutex_waiters()
				  if (wait_list_empty(l)) {
				    owner = l->owner & ~HAS_WAITERS;
cmpxchg(l->owner, T1, NULL)
 ===> Success (l->owner = NULL)
				    l->owner = owner
				     ==> l->owner = T1
				  }

That means the problem is caused by fixup_rt_mutex_waiters() which does the
RMW to clear the waiters bit unconditionally when there are no waiters in
the rtmutexes rbtree.

This can be fatal: A concurrent unlock can release the rtmutex in the
fastpath because the waiters bit is not set. If the cmpxchg() gets in the
middle of the RMW operation then the previous owner, which just unlocked
the rtmutex is set as the owner again when the write takes place after the
successfull cmpxchg().

The solution is rather trivial: verify that the owner member of the rtmutex
has the waiters bit set before clearing it. This does not require a
cmpxchg() or other atomic operations because the waiters bit can only be
set and cleared with the rtmutex wait_lock held. It's also safe against the
fast path unlock attempt. The unlock attempt via cmpxchg() will either see
the bit set and take the slowpath or see the bit cleared and release it
atomically in the fastpath.

It's remarkable that the test program provided by David triggers on ARM64
and MIPS64 really quick, but it refuses to reproduce on x86-64, while the
problem exists there as well. That refusal might explain that this got not
discovered earlier despite the bug existing from day one of the rtmutex
implementation more than 10 years ago.

Thanks to David for meticulously instrumenting the code and providing the
information which allowed to decode this subtle problem.

Reported-by: David Daney <ddaney@caviumnetworks.com>
Tested-by: David Daney <david.daney@cavium.com>
Signed-off-by: Thomas Gleixner <tglx@linutronix.de>
Reviewed-by: Steven Rostedt <rostedt@goodmis.org>
Acked-by: Peter Zijlstra (Intel) <peterz@infradead.org>
Cc: Linus Torvalds <torvalds@linux-foundation.org>
Cc: Mark Rutland <mark.rutland@arm.com>
Cc: Peter Zijlstra <peterz@infradead.org>
Cc: Sebastian Siewior <bigeasy@linutronix.de>
Cc: Will Deacon <will.deacon@arm.com>
Cc: stable@vger.kernel.org
Fixes: 23f78d4a ("[PATCH] pi-futex: rt mutex core")
Link: http://lkml.kernel.org/r/20161130210030.351136722@linutronix.de


Signed-off-by: Ingo Molnar <mingo@kernel.org>

parent 25139409

kernel/locking/rtmutex.c

+66 −2

Original line number	Diff line number	Diff line
		@@ -65,8 +65,72 @@ static inline void clear_rt_mutex_waiters(struct rt_mutex *lock)

		static void fixup_rt_mutex_waiters(struct rt_mutex *lock)
		{
		if (!rt_mutex_has_waiters(lock))
		clear_rt_mutex_waiters(lock);
		unsigned long owner, p = (unsigned long ) &lock->owner;

		if (rt_mutex_has_waiters(lock))
		return;

		/*
		* The rbtree has no waiters enqueued, now make sure that the
		* lock->owner still has the waiters bit set, otherwise the
		* following can happen:
		*
		* CPU 0 CPU 1 CPU2
		* l->owner=T1
		* rt_mutex_lock(l)
		* lock(l->lock)
		* l->owner = T1 \| HAS_WAITERS;
		* enqueue(T2)
		* boost()
		* unlock(l->lock)
		* block()
		*
		* rt_mutex_lock(l)
		* lock(l->lock)
		* l->owner = T1 \| HAS_WAITERS;
		* enqueue(T3)
		* boost()
		* unlock(l->lock)
		* block()
		* signal(->T2) signal(->T3)
		* lock(l->lock)
		* dequeue(T2)
		* deboost()
		* unlock(l->lock)
		* lock(l->lock)
		* dequeue(T3)
		* ==> wait list is empty
		* deboost()
		* unlock(l->lock)
		* lock(l->lock)
		* fixup_rt_mutex_waiters()
		* if (wait_list_empty(l) {
		* l->owner = owner
		* owner = l->owner & ~HAS_WAITERS;
		* ==> l->owner = T1
		* }
		* lock(l->lock)
		* rt_mutex_unlock(l) fixup_rt_mutex_waiters()
		* if (wait_list_empty(l) {
		* owner = l->owner & ~HAS_WAITERS;
		* cmpxchg(l->owner, T1, NULL)
		* ===> Success (l->owner = NULL)
		*
		* l->owner = owner
		* ==> l->owner = T1
		* }
		*
		* With the check for the waiter bit in place T3 on CPU2 will not
		* overwrite. All tasks fiddling with the waiters bit are
		* serialized by l->lock, so nothing else can modify the waiters
		* bit. If the bit is set then nothing can change l->owner either
		* so the simple RMW is safe. The cmpxchg() will simply fail if it
		* happens in the middle of the RMW because the waiters bit is
		* still set.
		*/
		owner = READ_ONCE(*p);
		if (owner & RT_MUTEX_HAS_WAITERS)
		WRITE_ONCE(*p, owner & ~RT_MUTEX_HAS_WAITERS);
		}

		/*

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