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Date: Mon, 22 Jan 2024 22:48:52 -0500
From: Steven Rostedt <rostedt@goodmis.org>
To: Yajun Deng <yajun.deng@linux.dev>
Cc: mingo@redhat.com, peterz@infradead.org, juri.lelli@redhat.com,
 vincent.guittot@linaro.org, dietmar.eggemann@arm.com, bsegall@google.com,
 mgorman@suse.de, bristot@redhat.com, vschneid@redhat.com,
 linux-kernel@vger.kernel.org
Subject: Re: [PATCH v2 1/2] sched/fair: Return NULL when entity isn't a task
 in task_of()
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In-Reply-To: <20231206063400.3206346-2-yajun.deng@linux.dev>
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	<20231206063400.3206346-2-yajun.deng@linux.dev>
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On Wed,  6 Dec 2023 14:33:59 +0800
Yajun Deng <yajun.deng@linux.dev> wrote:

> Before calling task_of(), we need to make sure that the entity is a task.
> There is also a warning in task_of() if the entity isn't a task. That
> means we need to check the entity twice. If the entity isn't a task,

Does it really check it twice? Have you disassembled it to see if the code
is any better?

#define entity_is_task(se)	(!se->my_q)
static inline struct task_struct *task_of(struct sched_entity *se)
{
	SCHED_WARN_ON(!entity_is_task(se));
	return container_of(se, struct task_struct, se);
}

The above is a macro and a static inline, which means that the compiler
should optimized out that second check.


> return the task struct is meaningless.
> 
> Return NULL when entity isn't a task in task_of(), and call task_of()
> instead of entity_is_task() when we need a task_struct.

I'm not against the change, as it could be considered a clean up. But it is
up to the sched maintainers to decide if it's worth the churn.

-- Steve


> 
> Signed-off-by: Yajun Deng <yajun.deng@linux.dev>