Re: Question on mutex code

From: Rabin Vincent
Date: Sun Mar 15 2015 - 18:10:36 EST


On Sun, Mar 15, 2015 at 11:49:07PM +0200, Matthias Bonne wrote:
> So both mutex_trylock() and mutex_unlock() always use the slow paths.
> The slowpath for mutex_unlock() is __mutex_unlock_slowpath(), which
> simply calls __mutex_unlock_common_slowpath(), and the latter starts
> like this:
>
> /*
> * As a performance measurement, release the lock before doing other
> * wakeup related duties to follow. This allows other tasks to
> acquire
> * the lock sooner, while still handling cleanups in past unlock
> calls.
> * This can be done as we do not enforce strict equivalence between
> the
> * mutex counter and wait_list.
> *
> *
> * Some architectures leave the lock unlocked in the fastpath
> failure
> * case, others need to leave it locked. In the later case we have
> to
> * unlock it here - as the lock counter is currently 0 or negative.
> */
> if (__mutex_slowpath_needs_to_unlock())
> atomic_set(&lock->count, 1);
>
> spin_lock_mutex(&lock->wait_lock, flags);
> [...]
>
> So the counter is set to 1 before taking the spinlock, which I think
> might cause the race. Did I miss something?

Yes, you miss the fact that __mutex_slowpath_needs_to_unlock() is 0 for
the CONFIG_DEBUG_MUTEXES case:

#ifdef CONFIG_DEBUG_MUTEXES
# include "mutex-debug.h"
# include <asm-generic/mutex-null.h>
/*
* Must be 0 for the debug case so we do not do the unlock outside of the
* wait_lock region. debug_mutex_unlock() will do the actual unlock in this
* case.
*/
# undef __mutex_slowpath_needs_to_unlock
# define __mutex_slowpath_needs_to_unlock() 0
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