forked from Minki/linux
rcu: Remove fast check path from __synchronize_srcu()
The fastpath in __synchronize_srcu() is designed to handle cases where there are a large number of concurrent calls for the same srcu_struct structure. However, the Linux kernel currently does not use SRCU in this manner, so remove the fastpath checks for simplicity. Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> Signed-off-by: Paul E. McKenney <paulmck@linux.vnet.ibm.com>
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@ -308,7 +308,7 @@ static void flip_idx_and_wait(struct srcu_struct *sp, bool expedited)
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*/
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static void __synchronize_srcu(struct srcu_struct *sp, bool expedited)
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{
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int idx;
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int idx = 0;
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rcu_lockdep_assert(!lock_is_held(&sp->dep_map) &&
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!lock_is_held(&rcu_bh_lock_map) &&
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@ -316,31 +316,8 @@ static void __synchronize_srcu(struct srcu_struct *sp, bool expedited)
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!lock_is_held(&rcu_sched_lock_map),
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"Illegal synchronize_srcu() in same-type SRCU (or RCU) read-side critical section");
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smp_mb(); /* Ensure prior action happens before grace period. */
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idx = ACCESS_ONCE(sp->completed);
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smp_mb(); /* Access to ->completed before lock acquisition. */
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mutex_lock(&sp->mutex);
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/*
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* Check to see if someone else did the work for us while we were
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* waiting to acquire the lock. We need -three- advances of
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* the counter, not just one. If there was but one, we might have
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* shown up -after- our helper's first synchronize_sched(), thus
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* having failed to prevent CPU-reordering races with concurrent
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* srcu_read_unlock()s on other CPUs (see comment below). If there
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* was only two, we are guaranteed to have waited through only one
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* full index-flip phase. So we either (1) wait for three or
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* (2) supply the additional ones we need.
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*/
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if (sp->completed == idx + 2)
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idx = 1;
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else if (sp->completed == idx + 3) {
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mutex_unlock(&sp->mutex);
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return;
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} else
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idx = 0;
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/*
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* If there were no helpers, then we need to do two flips of
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* the index. The first flip is required if there are any
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