Re: [PATCH v4] mm: fix is_pinnable_page against on cma page

John Hubbard
2022-05-17T20:12:14Z

On 5/17/22 12:28, Jason Gunthorpe wrote:
>> If you compare this to the snippet above, you'll see that there is
>> an extra mov statement, and that one dereferences a pointer from
>> %rax:
>>
>> mov (%rax),%rbx
>
> That is the same move as:
>
> mov 0x8(%rdx,%rax,8),%rbx
>
> Except that the EA calculation was done in advance and stored in rax.
>
> lea isn't a memory reference, it is just computing the pointer value
> that 0x8(%rdx,%rax,8) represents. ie the lea computes
>
> %rax = %rdx + %rax*8 + 6
>
> Which is then fed into the mov. Maybe it is an optimization to allow
> one pipe to do the shr and an other to the EA - IDK, seems like a
> random thing for the compiler to do.

Apologies for getting that wrong, and thanks for walking me through the
asm.

[...]
>
> Paul can correct me, but I understand we do not have a list of allowed
> operations that are exempted from the READ_ONCE() requirement. ie it
> is not just conditional branching that requires READ_ONCE().
>
> This is why READ_ONCE() must always be on the memory load, because the
> point is to sanitize away the uncertainty that comes with an unlocked
> read of unstable memory contents. READ_ONCE() samples the value in
> memory, and removes all tearing, multiload, etc "instability" that may
> effect down stream computations. In this way down stream compulations
> become reliable.
>
> Jason

So then:

diff --git a/mm/page_alloc.c b/mm/page_alloc.c
index 0e42038382c1..b404f87e2682 100644
--- a/mm/page_alloc.c
+++ b/mm/page_alloc.c
@@ -482,7 +482,12 @@ unsigned long __get_pfnblock_flags_mask(const struct page *page,
word_bitidx = bitidx / BITS_PER_LONG;
bitidx &= (BITS_PER_LONG-1);

- word = bitmap[word_bitidx];
+ /*
+ * This races, without locks, with set_pageblock_migratetype(). Ensure
+ * a consistent (non-tearing) read of the memory array, so that results,
+ * even though racy, are not corrupted.
+ */
+ word = READ_ONCE(bitmap[word_bitidx]);
return (word >> bitidx) & mask;
}

thanks,
--
John Hubbard
NVIDIA