Question on mutex code

[rabin@rab.in (Rabin Vincent)]

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