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Message-ID: <cbd68d63-ac48-7b36-d317-7bb2b480e6f7@huawei.com>
Date: Tue, 9 Jul 2019 09:25:45 +0800
From: Zhiqiang Liu <liuzhiqiang26@...wei.com>
To: Thomas Gleixner <tglx@...utronix.de>
CC: <corbet@....net>, <mcgrof@...nel.org>,
Kees Cook <keescook@...omium.org>, <akpm@...ux-foundation.org>,
<manfred@...orfullife.com>, <jwilk@...lk.net>,
<dvyukov@...gle.com>, <feng.tang@...el.com>,
<sunilmut@...rosoft.com>, <quentin.perret@....com>,
<linux@...mhuis.info>, <alex.popov@...ux.com>,
<linux-doc@...r.kernel.org>, <linux-kernel@...r.kernel.org>,
<linux-fsdevel@...r.kernel.org>,
"wangxiaogang (F)" <wangxiaogang3@...wei.com>,
"Zhoukang (A)" <zhoukang7@...wei.com>,
Mingfangsen <mingfangsen@...wei.com>, <tedheadster@...il.com>,
Eric Dumazet <edumazet@...gle.com>
Subject: Re: [PATCH next] softirq: enable MAX_SOFTIRQ_TIME tuning with sysctl
max_softirq_time_usecs
On 2019/7/8 22:14, Thomas Gleixner wrote:
> Zhiqiang,
>
>> If HZ satisfies the condition: HZ <= MSEC_PER_SEC && !(MSEC_PER_SEC %
>> HZ), the return value of _msecs_to_jiffies func with m=0 is different
>> with different HZ setting.
>
>> ------------------------------------
>> | HZ | MSEC_PER_SEC / HZ | return |
>> ------------------------------------
>> |1000| 1 | 0 |
>> |500 | 2 | 1 |
>> |200 | 5 | 1 |
>> |100 | 10 | 1 |
>> ------------------------------------
>>
>> Why only the return value of HZ=1000 is equal to 0 with m=0 ?
>
> I don't know how you tested that, but obviously all four HZ values use
> this variant:
>
>> #if HZ <= MSEC_PER_SEC && !(MSEC_PER_SEC % HZ)
>> static inline unsigned long _msecs_to_jiffies(const unsigned int m)
>> {
>> return (m + (MSEC_PER_SEC / HZ) - 1) / (MSEC_PER_SEC / HZ);
>> }
>
> and for all four HZ values the result is 0. Why?
>
> For m = 0 the calculation reduces to:
>
> ((MSEC_PER_SEC / HZ) - 1) / (MSEC_PER_SEC / HZ)
>
> i.e.
>
> (x - 1) / x where x = [1, 2, 5, 10]
>
> which is guaranteed to be 0 for integer math. If not, you have a compiler
> problem.
>
> Thanks,
>
> tglx
Thanks for your reply. Actually, I have made a low-level mistake.
I am really sorry for that.
Thanks again.
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