RE: [PATCH 10/10] powerpc: remove address space overrides using set_fs()

David Laight
2020-09-02T15:40:29Z

From: Christophe Leroy
> Sent: 02 September 2020 15:13
>
>
> Le 02/09/2020 à 15:51, David Laight a écrit :
> > From: Christophe Leroy
> >> Sent: 02 September 2020 14:25
> >> Le 02/09/2020 à 15:13, David Laight a écrit :
> >>> From: Christoph Hellwig
> >>>> Sent: 02 September 2020 13:37
> >>>>
> >>>> On Wed, Sep 02, 2020 at 08:15:12AM +0200, Christophe Leroy wrote:
> >>>>>> - return 0;
> >>>>>> - return (size == 0 || size - 1 <= seg.seg - addr);
> >>>>>> + if (addr >= TASK_SIZE_MAX)
> >>>>>> + return false;
> >>>>>> + if (size == 0)
> >>>>>> + return false;
> >>>>>
> >>>>> __access_ok() was returning true when size == 0 up to now. Any reason to
> >>>>> return false now ?
> >>>>
> >>>> No, this is accidental and broken. Can you re-run your benchmark with
> >>>> this fixed?
> >>>
> >>> Is TASK_SIZE_MASK defined such that you can do:
> >>>
> >>> return (addr | size) < TASK_SIZE_MAX) || !size;
> >>
> >> TASK_SIZE_MAX will usually be 0xc0000000
> >>
> >> With:
> >> addr = 0x80000000;
> >> size = 0x80000000;
> >>
> >> I expect it to fail ....
> >>
> >> With the formula you propose it will succeed, won't it ?
> >
> > Hmmm... Was i getting confused about some comments for 64bit
> > about there being such a big hole between valid user and kernel
> > addresses that it was enough to check that 'size < TASK_SIZE_MAX'.
> >
> > That would be true for 64bit x86 (and probably ppc (& arm??))
> > if TASK_SIZE_MAX were 0x4 << 60.
> > IIUC the highest user address is (much) less than 0x0 << 60
> > and the lowest kernel address (much) greater than 0xf << 60
> > on all these 64bit platforms.
> >
> > Actually if doing access_ok() inside get_user() you don't
> > need to check the size at all.
>
> You mean on 64 bit or on any platform ?

64bit and 32bit

> What about a word write to 0xbffffffe, won't it overwrite 0xc0000000 ?
>
> > You don't even need to in copy_to/from_user() provided
> > it always does a forwards copy.
>
> Do you mean due to the gap ?
> Is it garantied to be a gap ? Even on a 32 bits having TASK_SIZE set to
> 0xc0000000 and PAGE_OFFSET set to the same ?

I read somewhere (I won't find it again) that the last 4k page
(below 0xc0000000) must not be allocated on i386 because some
cpu (both intel and amd) do 'horrid things' if they try to
(IIRC) do instruction prefetches across the boundary.
So the accesses to 0xbffffffe will fault and the one to 0xc0000000
won't happen (in any useful way at least).

I'd suspect that not allocating the 3G-4k page would be a safe
bet on all architectures - even 68k.

David

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