Re: [PATCH 10/28] i386: map enough initial memory to create lowmem mappings

Eric W. Biederman
2007-04-23T09:14:58Z

Jeremy Fitzhardinge <jeremy@goop.org> writes:

> Chuck Ebbert wrote:
>> H. Peter Anvin wrote:
>>
>>> Andi Kleen wrote:
>>>
>>>> Then we would have seen reports surely?
>>>>
>
> Yes, I would have thought so. It surprised me that such an obvious bug
> could be there, apparently for a long time. But it's real, and
> potentially affects everyone. It probably doesn't affect highly modular
> distros much, since the kernel itself will be relatively small.
>
>> I never saw a description of the symptoms of encountering this bug.
>> Does it just hang, or what?
>>
>
> You get an early-fault message on-screen, assuming that's enabled;
> otherwise it will just appear to hang. It happens in pagetable_init,
> when it allocates a new pagetable above the head.S mapping (8M in my
> case). It will only hit if the kernel size approaches a 4M boundary,
> since it won't leave enough space mapped to construct the lowmem mappings.
>
> It only affects native booting, since under Xen all those mappings have
> already been constructed. It happened to me with a paravirt kernel that
> happened to Xen compiled into it, but that was irrelevent (though
> misleading; the 40k difference in kernel size was enough to make it not
> happen in a non-Xen kernel).

I happened to be looking at this stretch of code and I have realized
that this is quite simply the wrong fix.

The problem is that it depends intimately on the details of
alloc_bootmem_pages_low. Essentially the problem is that when
we are setting up the identity mappings in paging_init we assume
the identity mappings already exist.

If there are holes in the memory map or someone changes the way
pages are returned from alloc_bootmem_pages_low() this code
will break again.

The only way to ensure this will not happen is to do what we do
on x86_64 and map the new page table page into our address space
before we write to it. Assuming the page we allocate is already
mapped is simply not robust.

Eric