sched/swait: Remove __prepare_to_swait
There is no public user of this API, remove it. Signed-off-by: Peter Zijlstra (Intel) <peterz@infradead.org> Signed-off-by: Thomas Gleixner <tglx@linutronix.de> Acked-by: Linus Torvalds <torvalds@linux-foundation.org> Cc: bigeasy@linutronix.de Cc: oleg@redhat.com Cc: paulmck@linux.vnet.ibm.com Cc: pbonzini@redhat.com Link: https://lkml.kernel.org/r/20180612083909.157076812@infradead.org
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@ -161,7 +161,6 @@ extern void swake_up(struct swait_queue_head *q);
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extern void swake_up_all(struct swait_queue_head *q);
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extern void swake_up_locked(struct swait_queue_head *q);
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extern void __prepare_to_swait(struct swait_queue_head *q, struct swait_queue *wait);
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extern void prepare_to_swait(struct swait_queue_head *q, struct swait_queue *wait, int state);
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extern long prepare_to_swait_event(struct swait_queue_head *q, struct swait_queue *wait, int state);
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@ -69,7 +69,7 @@ void swake_up_all(struct swait_queue_head *q)
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}
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EXPORT_SYMBOL(swake_up_all);
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void __prepare_to_swait(struct swait_queue_head *q, struct swait_queue *wait)
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static void __prepare_to_swait(struct swait_queue_head *q, struct swait_queue *wait)
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{
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wait->task = current;
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if (list_empty(&wait->task_list))
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