Re: [PATCH 1/4] numa: introduce per-cgroup numa balancing locality, statistic

From: Peter Zijlstra <peterz@infradead.org>
To: 王贇 <yun.wang@linux.alibaba.com>
Cc: hannes@cmpxchg.org, mhocko@kernel.org, vdavydov.dev@gmail.com,
	Ingo Molnar <mingo@redhat.com>,
	linux-kernel@vger.kernel.org, linux-mm@kvack.org,
	mcgrof@kernel.org, keescook@chromium.org,
	linux-fsdevel@vger.kernel.org, cgroups@vger.kernel.org,
	Mel Gorman <mgorman@suse.de>,
	riel@surriel.com
Subject: Re: [PATCH 1/4] numa: introduce per-cgroup numa balancing locality, statistic
Date: Thu, 11 Jul 2019 15:47:54 +0200	[thread overview]
Message-ID: <20190711134754.GD3402@hirez.programming.kicks-ass.net> (raw)
In-Reply-To: <3ac9b43a-cc80-01be-0079-df008a71ce4b@linux.alibaba.com>

On Wed, Jul 03, 2019 at 11:28:10AM +0800, 王贇 wrote:

> @@ -3562,10 +3563,53 @@ static int memcg_numa_stat_show(struct seq_file *m, void *v)
>  		seq_putc(m, '\n');
>  	}
> 
> +#ifdef CONFIG_NUMA_BALANCING
> +	seq_puts(m, "locality");
> +	for (nr = 0; nr < NR_NL_INTERVAL; nr++) {
> +		int cpu;
> +		u64 sum = 0;
> +
> +		for_each_possible_cpu(cpu)
> +			sum += per_cpu(memcg->stat_numa->locality[nr], cpu);
> +
> +		seq_printf(m, " %u", jiffies_to_msecs(sum));
> +	}
> +	seq_putc(m, '\n');
> +#endif
> +
>  	return 0;
>  }
>  #endif /* CONFIG_NUMA */
> 
> +#ifdef CONFIG_NUMA_BALANCING
> +
> +void memcg_stat_numa_update(struct task_struct *p)
> +{
> +	struct mem_cgroup *memcg;
> +	unsigned long remote = p->numa_faults_locality[3];
> +	unsigned long local = p->numa_faults_locality[4];
> +	unsigned long idx = -1;
> +
> +	if (mem_cgroup_disabled())
> +		return;
> +
> +	if (remote || local) {
> +		idx = ((local * 10) / (remote + local)) - 2;
> +		/* 0~29% in one slot for cache align */
> +		if (idx < PERCENT_0_29)
> +			idx = PERCENT_0_29;
> +		else if (idx >= NR_NL_INTERVAL)
> +			idx = NR_NL_INTERVAL - 1;
> +	}
> +
> +	rcu_read_lock();
> +	memcg = mem_cgroup_from_task(p);
> +	if (idx != -1)
> +		this_cpu_inc(memcg->stat_numa->locality[idx]);

I thought cgroups were supposed to be hierarchical. That is, if we have:

          R
	 / \
	 A
	/\
	  B
	  \
	   t1

Then our task t1 should be accounted to B (as you do), but also to A and
R.

> +	rcu_read_unlock();
> +}
> +#endif