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>
This commit is contained in:
Matthew Wilcox (Oracle)
2019-07-04 15:13:23 -07:00
committed by Ingo Molnar
parent 3c29e651e1
commit 7b3c92b85a
5 changed files with 12 additions and 15 deletions

View File

@@ -628,8 +628,7 @@ static int rt_mutex_adjust_prio_chain(struct task_struct *task,
}
/* [10] Grab the next task, i.e. owner of @lock */
task = rt_mutex_owner(lock);
get_task_struct(task);
task = get_task_struct(rt_mutex_owner(lock));
raw_spin_lock(&task->pi_lock);
/*
@@ -709,8 +708,7 @@ static int rt_mutex_adjust_prio_chain(struct task_struct *task,
}
/* [10] Grab the next task, i.e. the owner of @lock */
task = rt_mutex_owner(lock);
get_task_struct(task);
task = get_task_struct(rt_mutex_owner(lock));
raw_spin_lock(&task->pi_lock);
/* [11] requeue the pi waiters if necessary */