[<prev] [next>] [<thread-prev] [thread-next>] [day] [month] [year] [list]
Message-ID: <53af9b5b-ac93-eaf9-8551-75fb25a243aa@huawei.com>
Date: Wed, 14 Jun 2017 12:35:07 +0100
From: John Garry <john.garry@...wei.com>
To: Will Deacon <will.deacon@....com>,
Mark Rutland <mark.rutland@....com>
CC: Shaokun Zhang <zhangshaokun@...ilicon.com>,
<linux-kernel@...r.kernel.org>,
<linux-arm-kernel@...ts.infradead.org>, <anurup.m@...wei.com>,
<tanxiaojun@...wei.com>, <xuwei5@...ilicon.com>,
<sanil.kumar@...ilicon.com>, <gabriele.paoloni@...wei.com>,
<shiju.jose@...wei.com>, <huangdaode@...ilicon.com>,
<linuxarm@...wei.com>, <dikshit.n@...wei.com>,
<shyju.pv@...wei.com>, <anurupvasu@...il.com>
Subject: Re: [PATCH v8 6/9] drivers: perf: hisi: Add support for Hisilicon
Djtag driver
On 14/06/2017 12:01, Will Deacon wrote:
> On Wed, Jun 14, 2017 at 11:42:30AM +0100, Mark Rutland wrote:
>> On Wed, Jun 14, 2017 at 11:06:58AM +0100, Will Deacon wrote:
>>> Apologies, I misunderstood your algorithm (I thought step (a) was on one CPU
>>> and step (b) was on another). Still, I don't understand the need for the
>>> timeout. If you instead read back the flag immediately, wouldn't it still
>>> work? e.g.
>>>
>>>
>>> lock:
>>> Readl_relaxed flag
>>> if (locked)
>>> goto lock;
>>>
>>> Writel_relaxed unique ID to flag
>>> Readl flag
>>> if (locked by somebody else)
>>> goto lock;
>>>
>>> <critical section>
>>>
>>> unlock:
>>> Writel unlocked value to flag
>>>
>>>
>>> Given that we're dealing with iomem, I think it will work, but I could be
>>> missing something obvious.
>>
>> Don't we have the race below where both threads can enter the critical
>> section?
>>
>> // flag f initial zero (unlocked)
>>
>> // t1, flag 1 // t2, flag 2
>> readl(f); // reads 0 l = readl(f); // reads 0
>>
>> <thinks lock is free> <thinks lock is free>
>>
>> writel(1, f);
>> readl(f); // reads 1
>> <thinks lock owned>
>> writel(2, f);
>> readl(f) // reads 2
>> <thinks lock owned>
>>
>> <crticial section> <critical section>
>
> Urgh, yeah, of course and *that's* what the udelay is trying to avoid,
> by "ensuring" that the <thinks lock is free> time and subsequent write
> propagation is all over before we re-read the flag.
>
> John -- how much space do you have on this device? Do you have, e.g. a byte
> for each CPU?
Hi Will,
To be clear, the agents in our case are the kernel and UEFI. Within the
kernel, we use a kernel spinlock to lock the same djtag between threads,
for these reasons:
- kernel has a native spinlock
- we are limited in locking values, as the lock flag is only a 8b field
in v2 hw (called module select)
Thanks,
John
>
> Will
>
> .
>
Powered by blists - more mailing lists