Date: Sun, 06 Mar 2005 11:21:46 +0800 From: David Xu <davidxu@freebsd.org> To: Sam Lawrance <boris@brooknet.com.au> Cc: current@freebsd.org Subject: Re: Swapped out procs not brought in immediately after child exits Message-ID: <422A774A.2070001@freebsd.org> In-Reply-To: <1110077369.790.48.camel@dirk.no.domain> References: <20050306012146.701FB17D8@localhost> <422A6046.5080801@freebsd.org> <1110077369.790.48.camel@dirk.no.domain>
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Sam Lawrance wrote: >On Sun, 2005-03-06 at 09:43 +0800, David Xu wrote: > > >>Sam Lawrance wrote: >> >> >> >>>>How-To-Repeat: >>>> >>>> >>>> >>>> >>>Run a shell somewhere (first). Su or run another shell or similar (second). >>>Wait until the first shell has swapped out (might require running some other >>>memory hogs). Exit the second shell. Notice that the second shell takes a >>>long time to exit. >>> >>> >>> >>> >>> >>This reminds me that it is another swappable kernel stack problem, if we >>don't have >>it, we even needn't TDP_WAKEPROC0 hack, interesting. :) >> >> > >Do I understand this correctly: When a process is swapped back in, the >kernel stack is faulted in immediately and user space is faulted in as >needed? > >And without swappable kernel stack, no extra action is required because >the kernel stack is already in, and user space will be faulted in as >usual? > > > Yes, you are right.
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