Messages in this thread | | | Date | Wed, 11 May 2022 17:49:49 -0700 | From | "Paul E. McKenney" <> | Subject | Re: [PATCH v4] mm: fix is_pinnable_page against on cma page |
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On Wed, May 11, 2022 at 05:34:52PM -0700, John Hubbard wrote: > On 5/11/22 17:26, Minchan Kim wrote: > > > > Let me try to say this more clearly: I don't think that the following > > > > __READ_ONCE() statement can actually help anything, given that > > > > get_pageblock_migratetype() is non-inlined: > > > > > > > > + int __mt = get_pageblock_migratetype(page); > > > > + int mt = __READ_ONCE(__mt); > > > > + > > > > + if (mt & (MIGRATE_CMA | MIGRATE_ISOLATE)) > > > > + return false; > > > > > > > > > > > > Am I missing anything here? > > > > > > In the absence of future aggression from link-time optimizations (LTO), > > > you are missing nothing. > > > > A thing I want to note is Android kernel uses LTO full mode. > > Thanks Paul for explaining the state of things. > > Minchan, how about something like very close to your original draft, > then, but with a little note, and the "&" as well: > > int __mt = get_pageblock_migratetype(page); > > /* > * Defend against future compiler LTO features, or code refactoring > * that inlines the above function, by forcing a single read. Because, this > * routine races with set_pageblock_migratetype(), and we want to avoid > * reading zero, when actually one or the other flags was set. > */ > int mt = __READ_ONCE(__mt); > > if (mt & (MIGRATE_CMA | MIGRATE_ISOLATE)) > return false; > > > ...which should make everyone comfortable and protected from the > future sins of the compiler and linker teams? :)
This would work, but it would force a store to the stack and an immediate reload. Which might be OK on this code path.
But using READ_ONCE() in (I think?) __get_pfnblock_flags_mask() would likely generate the same code that is produced today.
word = READ_ONCE(bitmap[word_bitidx]);
But I could easily have missed a turn in that cascade of functions. ;-)
Or there might be some code path that really hates a READ_ONCE() in that place.
Thanx, Paul
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