Re: GSoC ticket #1069 (thread scheduler) application

On 4/11/07, Ingo Weinhold <bonefish-CFLBMwTPW48UNGrzBIF7/[email protected]> wrote:
> Please note that: (1) There are usually not that many threads waiting for a
> single semaphore anyway. (2) The most likely semaphore release operation is
> with count 1, i.e. only the frontmost thread in the semaphore's queue will be
> woken up. (3) Even in cases were all waiting threads are woken up (e.g.
> delete_sem()) some info for each one needs to be updated, so the operation is
> "O(n)" anyway.

I might have a hidden ace on my sleeves... Who knows ;)

Could you please explain what info must be updated on every thread?
And if you have time to explain me some more kernel internals, please,
I'm online on #haiku. I promise I'll try not to ask way too many
stupid questions :)

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