What it will do is change entries on the page table. We have to be sure
two processes don't read/evict the same page in at the same time.
> The swapout code doesn't need to know one way or the other, because the
> swapout code never actually touches the pmd itself, it just follows the
> pointers - it doesn't ever need to worry about the pmd counts at all.
That was my original, incomplete view of the situation at first as well.
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