Re: Question on mutex code
From: Valdis . Kletnieks
Date: Tue Mar 10 2015 - 11:01:20 EST
On Tue, 10 Mar 2015 14:03:59 +0100, Yann Droneaud said:
> > Consider the following sequence of events:
> > 0. Suppose a mutex is locked by task A and has no waiters.
> > 1. Task B calls mutex_trylock().
> > 2. mutex_trylock() calls the architecture-specific
> > __mutex_fastpath_trylock(), with __mutex_trylock_slowpath() as
> > fail_fn.
> > 3. According to the description of __mutex_fastpath_trylock() (for
> > example in include/asm-generic/mutex-dec.h), "if the architecture
> > has no effective trylock variant, it should call the fail_fn
> > spinlock-based trylock variant unconditionally". So
> > __mutex_fastpath_trylock() may now call __mutex_trylock_slowpath().
> > 4. Task A releases the mutex.
> > 5. Task B, in __mutex_trylock_slowpath, executes:
> > /* No need to trylock if the mutex is locked. */
> > if (mutex_is_locked(lock))
> > return 0;
> > Since the mutex is no longer locked, the function continues.
> > 6. Task C, which runs on a different cpu than task B, locks the mutex
> > again.
> > 7. Task B, in __mutex_trylock_slowpath(), continues:
> > spin_lock_mutex(&lock->wait_lock, flags);
B will spin here until C releases the lock.
When that spin exits, C no longer holds the lock. Re-do the analysis
from this point.
Description: PGP signature