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Date: Sat, 13 Feb 2010 00:09:23 +0000
From: Alan Cox <alan@lxorguk.ukuu.org.uk>
To: Henry Nestler <henry-20071123@arcor.de>
Cc: linux-kernel@vger.kernel.org,
       "=?ISO-8859-14?B?RnLpZOlyaWM=?= L. W. Meunier" <fredlwm@gmail.com>,
       gregkh@suse.de, Simon.Richter@phobos.fachschaften.tu-muenchen.de
Subject: Re: [2.6.33-rc7] tty: page allocation failure (from
 tty_buffer_request_room)
Message-ID: <20100213000923.03007ddf@lxorguk.ukuu.org.uk>
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References: <ed7v4-5Xl-13@gated-at.bofh.it>
	<4B75E528.7090605@arcor.de>
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> > You are doing some output on console, for example write to a device /dev/tty.
> > Inside tty_buffer_request_room I found, that kmalloc was called and interrupts
> > are disabled. That means, memory can not swap out - if no free page is
> > available. tty_buffer_request_room checks for NULL returns. So, it assumes,
> > that kmalloc can fail. Unless they have forgotten to set __GFP_NOWARN. - Or it
> > is a defective design, and kmalloc should not called inside a spin_lock.


Neither - if we can't allocate a small buffer (and I would expect this to
be a small buffer of under one page) then we lose data and while we
continue to run it indicates a very serious low memory condition that the
user really ought to be aware of.

If anything what this says is that tty_insert_flip_string and friends
ought to be tweaked to do

nt tty_insert_flip_string(struct tty_struct *tty, const unsigned char
*chars, size_t size)
{  
        int copied = 0;
        do {
		int want = min(size - copied, TTY_BUFFER_PAGE);
                int space = tty_buffer_request_room(tty, want);
		...


where TTY_BUFFER_PAGE is defined to be the size to request so the buffer
will fit in one page of memory.