Date: Sun, 24 Nov 2013 23:25:14 +0000 (UTC) From: Glen Barber <gjb@FreeBSD.org> To: doc-committers@freebsd.org, svn-doc-all@freebsd.org, svn-doc-head@freebsd.org Subject: svn commit: r43237 - head/en_US.ISO8859-1/books/faq Message-ID: <201311242325.rAONPEH6063596@svn.freebsd.org>
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Author: gjb Date: Sun Nov 24 23:25:14 2013 New Revision: 43237 URL: http://svnweb.freebsd.org/changeset/doc/43237 Log: Fix build. Sponsored by: The FreeBSD Foundation Modified: head/en_US.ISO8859-1/books/faq/book.xml Modified: head/en_US.ISO8859-1/books/faq/book.xml ============================================================================== --- head/en_US.ISO8859-1/books/faq/book.xml Sun Nov 24 22:44:32 2013 (r43236) +++ head/en_US.ISO8859-1/books/faq/book.xml Sun Nov 24 23:25:14 2013 (r43237) @@ -1339,22 +1339,22 @@ <answer> <para>For FFS file systems, the largest file system is practically - limited by the amount of memory required to &man.fsck.8 the file - system. &man.fsck.8 requires one bit per fragment, which with - the default fragment size of 4&nbps;KB equates to 32&nbps;MB + limited by the amount of memory required to &man.fsck.8; the file + system. &man.fsck.8; requires one bit per fragment, which with + the default fragment size of 4 KB equates to 32 MB of memory per TB of disk. This does mean that on architectures - which limit userland processes to 2&nbps;GB (e.g., &i386;), - the maximum &man.fsck.8'able filesystem is ~60&nbps;TB.</para> + which limit userland processes to 2 GB (e.g., &i386;), + the maximum &man.fsck.8;'able filesystem is ~60 TB.</para> - <para>If there was not a &man.fsck.8 memory limit the maximum - filesystem size would be 2&nbps;^&nbps;64 (blocks) * 32&nbps;KB - => 16 Exa * 32&nbps;KB => 512 ZettaBytes.</para> + <para>If there was not a &man.fsck.8; memory limit the maximum + filesystem size would be 2 ^ 64 (blocks) * 32 KB + => 16 Exa * 32 KB => 512 ZettaBytes.</para> <para>The maximum size of a single FFS file is approximately 2 PB with the default block size of 32 KB. Each 32 KB block can point to 4096 blocks. With triple - indirect blocks, the calculation is 32&nbps;KB * 12 + - 32&nbps;KB * 4096 + 32&nbps;KB * 4096^2 + 32&nbps;KB * + indirect blocks, the calculation is 32 KB * 12 + + 32 KB * 4096 + 32 KB * 4096^2 + 32 KB * 4096^3. Increasing the block size to 64 KB will increase the max file size by a factor of 16.</para> </answer>
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