linux-kernel - Re: [RFC PATCH 1/5] x86: introduce preemption disable prefix

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Date:   Fri, 19 Oct 2018 12:38:10 +0200
From:   Oleg Nesterov <oleg@...hat.com>
To:     Andy Lutomirski <luto@...nel.org>
Cc:     Nadav Amit <namit@...are.com>,
        Alexei Starovoitov <alexei.starovoitov@...il.com>,
        Ingo Molnar <mingo@...hat.com>,
        Peter Zijlstra <peterz@...radead.org>,
        "H. Peter Anvin" <hpa@...or.com>,
        Thomas Gleixner <tglx@...utronix.de>,
        LKML <linux-kernel@...r.kernel.org>, X86 ML <x86@...nel.org>,
        Borislav Petkov <bp@...en8.de>,
        "Woodhouse, David" <dwmw@...zon.co.uk>
Subject: Re: [RFC PATCH 1/5] x86: introduce preemption disable prefix

On 10/18, Andy Lutomirski wrote:
>
> Oleg, the code in kernel/signal.c:
>
>                 preempt_disable();
>                 read_unlock(&tasklist_lock);
>                 preempt_enable_no_resched();
>                 freezable_schedule();
>
> looks bogus.  I don't get what it's trying to achieve with
> preempt_disable(), and I also don't see why no_resched does anything.
> Sure, it prevents a reschedule triggered during read_unlock() from
> causing a reschedule,

Yes. Lets suppose we remove preempt_disable/enable.

Debugger was already woken up, if it runs on the same CPU quite possibly
it will preemt the tracee. After that debugger will spin in wait_task_inactive(),
until it is in turn preempted or calls schedule_timeout(1), so that the tracee
(current) can finally call __schedule(preempt = F) and call deactivate_task() to
become inactive.

> but it doesn't prevent an interrupt immediately
> after the preempt_enable_no_resched() call from scheduling.

Yes, but this is less likely.

Oleg.