Date: Tue, 30 Sep 2014 16:36:07 +0400 From: Gleb Smirnoff <glebius@FreeBSD.org> To: Andriy Gapon <avg@FreeBSD.org> Cc: freebsd-hackers@FreeBSD.org Subject: Re: uk_slabsize, uk_ppera, uk_ipers, uk_pages Message-ID: <20140930123607.GE73266@glebius.int.ru> In-Reply-To: <542A9EF0.3050405@FreeBSD.org> References: <542A916A.2060703@FreeBSD.org> <20140930114424.GD73266@glebius.int.ru> <542A9EF0.3050405@FreeBSD.org>
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On Tue, Sep 30, 2014 at 03:15:44PM +0300, Andriy Gapon wrote: A> This is not true for kegs with multi-page slabs. Consider a zone with 8KB items A> on a system 4KB pages. Its keg uses slabs with the size of two pages, uk_ppera A> is 2. There is only one item per slab, uk_ipers is 1. Let's say there are two A> slabs allocated. Then uk_pages is 4. So, uk_ipers * uk_pages would give 4, but A> in reality there are only two items. The correct calculation must be (uk_pages A> / uk_ppera) * uk_ipers. A> A> If you have enough CPUs for a pcpu zone to use multi-page slabs / allocations, A> then the above will also be applicable. Consider "64 pcpu" and 8 CPUs. You have A> uk_ppera = 2, uk_ipers = 128. If there is only 1 "real" slab allocated that's 2 A> pages, so uk_pages * uk_ipers = 256, but in reality the correct number of A> provided items is (uk_pages / uk_ppera) * uk_ipers = 128. You are right. -- Totus tuus, Glebius.
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