On Mon, 28 Sep 2009, Linus Torvalds wrote:
>
> Or did I miss something?
Never mind, it's all fine. It just depends on what you define "unlocked"
to be, and "old value matches high bits is unlocked" is obviously the more
attractive case than the one I was thinking of ("new value matches high
bits"), since it makes a zero initializer be unlocked.
And it's what x86 does too. I don't know why I expected the "new value
is the ticket" case.
Linus
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