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Date: Wed, 4 Aug 2010 11:02:48 +0900
From: KAMEZAWA Hiroyuki <kamezawa.hiroyu@...fujitsu.com>
To: Nigel Cunningham <nigel@...onice.net>
Cc: Bojan Smojver <bojan@...ursive.com>, linux-kernel@...r.kernel.org
Subject: Re: [PATCH]: Compress hibernation image with LZO (in-kernel)
On Wed, 04 Aug 2010 11:50:41 +1000
Nigel Cunningham <nigel@...onice.net> wrote:
> Hi.
>
> Sorry for the delay in replying.
>
> On 03/08/10 11:59, Bojan Smojver wrote:
> >> From all this, I only got "In swsusp_free()" printed on resume. So, it
> > seems that save_image() does indeed free those vmalloc()-ed buffers and
> > they are not saved in the image.
> >
> > I even put this in hibernate.c:
> > ---------------------
> > /* Restore control flow magically appears here */
> > restore_processor_state();
> > if (!in_suspend)
> > platform_leave(platform_mode);
> >
> > printk(KERN_ERR "Resumed, checking swsusp_lzo_buffers.\n");
> > if (swsusp_lzo_buffers) {
> > printk (KERN_ERR "Setting vmalloc() buffers.\n");
> > memset(swsusp_lzo_buffers, 0, 80 * PAGE_SIZE);
> > }
> > ---------------------
> >
> > This printed just "Resumed, checking swsusp_lzo_buffers.", meaning it
> > was already set to NULL.
> >
> > Any further comments on this? Nigel, what do you reckon?
>
> I don't see what all the fuss was about. save_image is called after the
> snapshot is made (hibernate called hibernation_snapshot to create the
> image, then swsusp_write which in turn calls save_image), so there's no
> possibility of the memory allocated by it being included in the image or
> of a memory leak ocuring as a result.
>
I'm sorry if I'm wrong. Here is logic in my mind. (I'm reading hibernate(void).
not sure about user.c )
(linenumber is from 2.6.34 global, plz ignore)
624 error = hibernation_snapshot(hibernation_mode == HIBERNATION_PLATFORM);
625 if (error)
626 goto Thaw;
627
628 if (in_suspend) {
629 unsigned int flags = 0;
630
631 if (hibernation_mode == HIBERNATION_PLATFORM)
632 flags |= SF_PLATFORM_MODE;
633 pr_debug("PM: writing image.\n");
634 error = swsusp_write(flags);
Then, swsusp_write() actually writes image to the disk. Right ?
It finally calls save_image(). save_image() executes following loop.
433 while (1) {
434 ret = snapshot_read_next(snapshot, PAGE_SIZE);
435 if (ret <= 0)
436 break;
437 ret = swap_write_page(handle, data_of(*snapshot), &bio);
438 if (ret)
439 break;
440 if (!(nr_pages % m))
441 printk(KERN_CONT "\b\b\b\b%3d%%", nr_pages / m);
442 nr_pages++;
443 }
Ok, here, the image written to disk is got by snapshot_read_next().
Then, what it does ?
==
1650 page = pfn_to_page(memory_bm_next_pfn(©_bm));
1651 if (PageHighMem(page)) {
1652 /* Highmem pages are copied to the buffer,
1653 * because we can't return with a kmapped
1654 * highmem page (we may not be called again).
1655 */
1656 void *kaddr;
1657
1658 kaddr = kmap_atomic(page, KM_USER0);
1659 memcpy(buffer, kaddr, PAGE_SIZE);
1660 kunmap_atomic(kaddr, KM_USER0);
1661 handle->buffer = buffer;
1662 } else {
1663 handle->buffer = page_address(page);
1664 }
==
Ok, what actually written to disk is an image of memory at save_image() is
called.
What happens at vmalloc ?
1. page tabels for vmalloc() is set up.
2. vm_struct object is allocated.
3. vmap is allcoated.
Above "3" states are all saved into the disk image, as "used".
Then, after resume, all vmalloc() area is resumed as "allocated".
Wrong ?
Thanks,
-Kame
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