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sched/core: Convert get_task_struct() to return the task
Returning the pointer that was passed in allows us to write slightly more idiomatic code. Convert a few users. Signed-off-by: Matthew Wilcox (Oracle) <willy@infradead.org> Signed-off-by: Peter Zijlstra (Intel) <peterz@infradead.org> Cc: Linus Torvalds <torvalds@linux-foundation.org> Cc: Peter Zijlstra <peterz@infradead.org> Cc: Thomas Gleixner <tglx@linutronix.de> Link: https://lkml.kernel.org/r/20190704221323.24290-1-willy@infradead.org Signed-off-by: Ingo Molnar <mingo@kernel.org>
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5 changed files with 12 additions and 15 deletions
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@ -628,8 +628,7 @@ static int rt_mutex_adjust_prio_chain(struct task_struct *task,
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}
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/* [10] Grab the next task, i.e. owner of @lock */
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task = rt_mutex_owner(lock);
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get_task_struct(task);
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task = get_task_struct(rt_mutex_owner(lock));
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raw_spin_lock(&task->pi_lock);
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/*
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@ -709,8 +708,7 @@ static int rt_mutex_adjust_prio_chain(struct task_struct *task,
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}
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/* [10] Grab the next task, i.e. the owner of @lock */
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task = rt_mutex_owner(lock);
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get_task_struct(task);
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task = get_task_struct(rt_mutex_owner(lock));
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raw_spin_lock(&task->pi_lock);
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/* [11] requeue the pi waiters if necessary */
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