Date: Wed, 19 Oct 2011 20:37:13 +0000 From: Alexander Best <arundel@freebsd.org> To: Chuck Swiger <cswiger@mac.com> Cc: freebsd-questions@freebsd.org Subject: Re: small du(1) question Message-ID: <20111019203713.GA19350@freebsd.org> In-Reply-To: <6620A8F5-523D-4A1E-9CC1-9C2D917BF0C2@mac.com> References: <20111019193413.GA9065@freebsd.org> <6620A8F5-523D-4A1E-9CC1-9C2D917BF0C2@mac.com>
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On Wed Oct 19 11, Chuck Swiger wrote: > On Oct 19, 2011, at 12:34 PM, Alexander Best wrote: > > the du(1) man page states the following: > > > > " > > -B blocksize > > Calculate block counts in blocksize byte blocks. This is differ- > > ent from the -k, -m options or setting BLOCKSIZE and gives an > > estimate of how much space the examined file hierarchy would > > require on a filesystem with the given blocksize. Unless in -A > > mode, blocksize is rounded up to the next multiple of 512. > > " > > > > is this a doc bug, or does du(1) really always assume that every filesystem's > > blocksize == 512? > > The default blocksize is 512 bytes. > > The -B option flag lets you tell du to assume a different filesystem blocksize. so when running freebsd on a hdd with a blocksize of 4k, a simple 'du -h' will always display incorrect results, unless '-B 4096' was also specified? isn't there a way to automatically query the blocksize of the underlying device, instead of always asuming the blocksize is 512 byte? cheers. alex > > Regards, > -- > -Chuck >
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