linux-kernel - Re: [patch 11/17] Implement immediate update via stop_machine

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Message-ID: <47FEEE0E.6020206@gmail.com>
Date:	Fri, 11 Apr 2008 13:50:22 +0900
From:	KOSAKI Motohiro <kosaki.motohiro@...il.com>
To:	Mathieu Desnoyers <mathieu.desnoyers@...ymtl.ca>
CC:	KOSAKI Motohiro <kosaki.motohiro@...fujitsu.com>,
	akpm@...ux-foundation.org, Ingo Molnar <mingo@...e.hu>,
	linux-kernel@...r.kernel.org, Andi Kleen <andi@...stfloor.org>,
	Rusty Russell <rusty@...tcorp.com.au>,
	Jason Baron <jbaron@...hat.com>, Adrian Bunk <bunk@...sta.de>,
	Alexey Dobriyan <adobriyan@...il.com>,
	Christoph Hellwig <hch@...radead.org>, akpm@...l.org,
	linuxdev-kanex@...css.fujitsu.com
Subject: Re: [patch 11/17] Implement immediate update via stop_machine_run

Hi  Mathieu,
>> it seems this patch is must, Why do you separate patch [10/17] and [11/17]?
>> this patch remove almost portion of [10/17].
>> IMHO these patch merge into 1 patch is better.
>>
>>     
>
> Hi Kosaki,
>
> You are right, I will merge them and resend them in the following post.
>   
Thanks

>>> +static int stop_machine_imv_update(void *imv_ptr)
>>> +{
>>> +	struct __imv *imv = imv_ptr;
>>> +
>>> +	if (!wrote_text) {
>>>       
>> it seems racy.
>> Why don't need test_and_set?
>>
>> I think your stop_machine_run(ALL_CPUS) call fn concurrency...
>>     
> The answer to this mistery is in include/linux/stop_machine.h modified
> by add-all-cpus-option-to-stop-machine-run.patch :
>
>
>  /**
>   * stop_machine_run: freeze the machine on all CPUs and run this function
>   * @fn: the function to run
>   * @data: the data ptr for the @fn()
> - * @cpu: the cpu to run @fn() on (or any, if @cpu == NR_CPUS.
> + * @cpu: if @cpu == n, run @fn() on cpu n
> + *       if @cpu == NR_CPUS, run @fn() on any cpu
> + *       if @cpu == ALL_CPUS, run @fn() first on the calling cpu, and then
> + *       concurrently on all the other cpus
>   *
>   * Description: This causes a thread to be scheduled on every other cpu,
>   * each of which disables interrupts, and finally interrupts are disabled
>
> Therefore, the first execution of the function is done before all other
> executions.
>
>   
Ah, OK. I understand.
Thanks.


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