linux-kernel - Re: linux.git: printk() problem

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Message-ID: <CA+55aFzW5YEXBy3hyrML5P87_FmQ_MDCDAodDwge6zwOOp9zkw@mail.gmail.com>
Date:   Mon, 24 Oct 2016 21:15:41 -0700
From:   Linus Torvalds <torvalds@...ux-foundation.org>
To:     Sergey Senozhatsky <sergey.senozhatsky.work@...il.com>
Cc:     Sergey Senozhatsky <sergey.senozhatsky@...il.com>,
        Joe Perches <joe@...ches.com>,
        Geert Uytterhoeven <geert@...ux-m68k.org>,
        Tetsuo Handa <penguin-kernel@...ove.sakura.ne.jp>,
        Linux Kernel Mailing List <linux-kernel@...r.kernel.org>,
        Petr Mladek <pmladek@...e.cz>, Tejun Heo <tj@...nel.org>,
        Calvin Owens <calvinowens@...com>,
        Steven Rostedt <rostedt@...dmis.org>,
        Andrew Morton <akpm@...ux-foundation.org>
Subject: Re: linux.git: printk() problem

On Mon, Oct 24, 2016 at 9:06 PM, Sergey Senozhatsky
<sergey.senozhatsky.work@...il.com> wrote:
>
> 1) the way we dumpstack on x86 (at least on x86) is a spaghetti of
> printk() and pr_cont() calls. for instance, arch/x86/kernel/dumpstack_64.c
> show_regs() does pr_cont() to print out the registers, while the stack and
> backtrace are printed with printk(). so, I assume, the backtrace now will
> look a bit upside-down, because cont lines are printed with the delay.
> correct?

No. Most cont lines never hit the delay, because when the line is
completed, it is flushed (and then printed synchronously, assuming it
can get the console lock).

So the timeout only ever comes into effect if the line isn't completed
in time at all. Which is actually very rare, and never happens for the
"let's print things out in multiple chinks because we're using a
loop".

Similarly, if a new printk() happens due to interleaving, the previous
buffered line is always flushed first, so buffering never causes
out-of-order behavior.

Basically, the only time the timer actually does anything is if
something just does a printk() without a newline, and no other
printouts happen for the next 0.1s.

> 2) flush on oops.

Again, really not an issue for the exact reason above: nothing is ever
buffered when something new is printed.

And all you need to guarantee that last line of the oops itself is
printed is that it has a newline.

So again, the timer only matters for the exceptional case, not for the
normal situation. It's literally there to guarantee basic timeliness.

         Linus