Re: [NT] Quantum of interrupted process
From: Maxim S. Shatskih (maxim_at_storagecraft.com)
Date: 05/16/04
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Date: Mon, 17 May 2004 02:08:33 +0400
From what I know, the quantum is accounted since the thread was context
switched to, so, the counting is reset on each context switch.
ISRs or kernel-user transition do not cause context switches, and thus do
not reset the quantum.
-- Maxim Shatskih, Windows DDK MVP StorageCraft Corporation maxim@storagecraft.com http://www.storagecraft.com "G. Turgij" <t0urguy@hotmail.com> wrote in message news:c868tj$2kc$01$2@news.t-online.com... > Hi there! > > Let's say we have three processes: P1, P2 and P3 > They are scheduled to run in this order: P1 -> P2 -> P3 > Execution is running in P1 and its quantum (Q1) is almost > over. > > Now P1 is interrupted and control is passed to a KMD while IRQL > changes to DISPATCH_LEVEL (or let's say a spinlock is active)... > > Now my question: > Will Q1 be frozen and continued from the point when the interrupt > occured, even if the kernel-mode code takes very long to execute? > > In other words: If Q1 and even Q2 would theoretically have passed > during the kernel operation, will the scheduler switch to P3 with Q3 > set to what's left for P3 - OR - does it return to P1 with the original > (frozen) Q1? > > > Thanks in advance! > > > Gregor Turgij < > >
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