On Tue, 19 May 2015, Davidlohr Bueso wrote:
>
> +/*
> + * Lockless alternative to rt_mutex_has_waiters() as we do not need the
> + * wait_lock to check if we are in, for instance, a transitional state
> + * after calling mark_rt_mutex_waiters().
Before I get into a state of brain melt, could you please explain that
in an understandable way?
rt_mutex_has_waiters() looks at the root pointer of the rbtree head
whether that's empty. You can do a lockless check of that as well,
right? So what's the FAST part of that function and how is that
related to a point after we called mark_rt_mutex_waiters()?
> + */
> +static inline bool rt_mutex_has_waiters_fast(struct rt_mutex *lock)
> +{
> + unsigned long val = (unsigned long)lock->owner;
> +
> + if (!val)
> + return false;
> + return val & RT_MUTEX_HAS_WAITERS;
> +}
> +
> +/*
> + * Initial check for entering the mutex spinning loop
> + */
> +static inline bool rt_mutex_can_spin_on_owner(struct rt_mutex *lock)
> +{
> + struct task_struct *owner;
> + /* default return to spin: if no owner, the lock is free */
Rather than having a comment in the middle of the variable declaration
section, I'd prefer a comment explaing the whole logic of this
function.
> + int ret = true;
> +static bool rt_mutex_optimistic_spin(struct rt_mutex *lock)
> +{
> + bool taken = false;
> +
> + preempt_disable();
> +
> + if (!rt_mutex_can_spin_on_owner(lock))
> + goto done;
> + /*
> + * In order to avoid a stampede of mutex spinners trying to
> + * acquire the mutex all at once, the spinners need to take a
> + * MCS (queued) lock first before spinning on the owner field.
> + */
> + if (!osq_lock(&lock->osq))
> + goto done;
Hmm. The queue lock is serializing potential spinners, right?
So that's going to lead to a potential priority ordering problem
because if a lower prio task wins the racing to the ocq_lock queue,
then the higher prio waiter will be queued behind and blocked from
taking the lock first.
Thanks,
tglx
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