linux-kernel - Re: [PATCH 13/32] Documentation, x86: Documentation for Intel resource allocation user interface

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Date:	Tue, 19 Jul 2016 14:32:12 +0200 (CEST)
From:	Thomas Gleixner <tglx@...utronix.de>
To:	"Luck, Tony" <tony.luck@...el.com>
cc:	Fenghua Yu <fenghua.yu@...el.com>, Ingo Molnar <mingo@...e.hu>,
	"H. Peter Anvin" <h.peter.anvin@...el.com>,
	Tejun Heo <tj@...nel.org>, Borislav Petkov <bp@...e.de>,
	Stephane Eranian <eranian@...gle.com>,
	Peter Zijlstra <peterz@...radead.org>,
	Marcelo Tosatti <mtosatti@...hat.com>,
	David Carrillo-Cisneros <davidcc@...gle.com>,
	Ravi V Shankar <ravi.v.shankar@...el.com>,
	Vikas Shivappa <vikas.shivappa@...ux.intel.com>,
	Sai Prakhya <sai.praneeth.prakhya@...el.com>,
	linux-kernel <linux-kernel@...r.kernel.org>, x86 <x86@...nel.org>
Subject: Re: [PATCH 13/32] Documentation, x86: Documentation for Intel resource
 allocation user interface

On Thu, 14 Jul 2016, Luck, Tony wrote:
> So the core part of __intel_rdt_sched_in() will look like:
> 
> 	/*
> 	 * Precedence rules:
> 	 * Processes assigned to an rdtgroup use that group
> 	 * wherever they run. If they don't have an rdtgroup
> 	 * we see if the current cpu has one and use it.
> 	 * If no specific rdtgroup was provided, we use the
> 	 * root_rdtgroup
> 	 */
> 	rdtgrp = current->rdtgroup;
> 	if (!rdtgrp) {
> 		rdtgrp = per_cpu(cpu_rdtgroup, cpu);
> 		if (!rdtgrp)
> 			rdtgrp = root_rdtgroup;
> 	}

That can be done simpler. The default cpu_rdtgroup should be root_rdtgroup. So
you spare one conditional.

Thanks

	tglx