* Thomas Gleixner <tglx@xxxxxxxxxxxxx> wrote:
> So the problem we need to solve is:
>
> retry:
> lock(B);
> if (!try_lock(A)) {
> unlock(B);
> cpu_relax();
> goto retry;
> }
>
> So instead of doing that proposed magic boost, we can do something
> more straight forward:
>
> retry:
> lock(B);
> if (!try_lock(A)) {
> lock_and_drop(A, B);
> unlock(A);
> goto retry;
> }
>
> lock_and_drop() queues the task as a waiter on A, drops B and then
> does the PI adjustment on A.
>
> Thoughts?
So why not do:
lock(B);
if (!trylock(A)) {
unlock(B);
lock(A);
lock(B);
}
?
Or, if this can be done, why didn't we do:
lock(A);
lock(B);
to begin with?
i.e. I'm not sure the problem is properly specified.
Thanks,
Ingo
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