linux-kernel - Re: [PATCH v4] x86,sched: allow topologies where NUMA nodes share an LLC

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Date:   Wed, 4 Apr 2018 11:42:11 -0700
From:   Tim Chen <tim.c.chen@...ux.intel.com>
To:     Alison Schofield <alison.schofield@...el.com>
Cc:     Thomas Gleixner <tglx@...utronix.de>,
        Ingo Molnar <mingo@...nel.org>,
        Dave Hansen <dave.hansen@...ux.intel.com>,
        Tony Luck <tony.luck@...el.com>,
        "H. Peter Anvin" <hpa@...ux.intel.com>,
        Borislav Petkov <bp@...en8.de>,
        Peter Zijlstra <peterz@...radead.org>,
        David Rientjes <rientjes@...gle.com>,
        Igor Mammedov <imammedo@...hat.com>,
        Prarit Bhargava <prarit@...hat.com>, brice.goglin@...il.com,
        x86@...nel.org, linux-kernel@...r.kernel.org
Subject: Re: [PATCH v4] x86,sched: allow topologies where NUMA nodes share an
 LLC

On 04/04/2018 10:38 AM, Alison Schofield wrote:
> On Wed, Apr 04, 2018 at 10:24:49AM -0700, Tim Chen wrote:
>> On 04/03/2018 02:12 PM, Alison Schofield wrote:
>>
>>> +
>>> +	/*
>>> +	 * topology_sane() considers LLCs that span NUMA nodes to be
>>> +	 * insane and will display a warning message. Bypass the call
>>> +	 * to topology_sane() for snc_cpu's to avoid that warning.
>>> +	 */
>>> +
>>> +	if (!topology_same_node(c, o) && x86_match_cpu(snc_cpu)) {
>>> +		/* Indicate that package has NUMA nodes inside: */
>>> +		x86_has_numa_in_package = true;
>>
>> Why does the x86_has_numa_in_package has to be set here when it would have
>> been done later in set_cpu_sibling_map?
> 
> Tim,
> I had that same thought when you commented on it previously. After 
> discussing w DaveH, decided that match_llc() and match_die(c,0)
> could be different and chose to be (cautiously) redundant.
> alisons

If it is redundant, I suggest it be removed, and only added if
there is truly a case where the current logic 

                if (match_die(c, o) && !topology_same_node(c, o))
                        x86_has_numa_in_package = true;

fails.  And also the modification of this logic should be at the
same place for easy code maintenance. 

Tim  

> 
> 
> 
>>
>>> +
>>> +		/*
>>> +		 * false means 'c' does not share the LLC of 'o'.
>>> +		 * Note: this decision gets reflected all the way
>>> +		 * out to userspace.
>>> +		 */
>>> +
>>> +		return false;
>>
>> Thanks.
>>
>> Tim