Date: Sat, 29 Aug 1998 15:06:39 -0400 (EDT) From: zhihuizhang <bf20761@binghamton.edu> To: hackers <freebsd-hackers@FreeBSD.ORG> Subject: Macro cbtorpos() in fs.h Message-ID: <Pine.SOL.L3.93.980829145113.18319A-100000@bingsun2>
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I am trying to understand the macro cbtorpos() defined below:
#define cbtorpos(fs, bno) \
(((bno) * NSPF(fs) % (fs)->fs_spc / (fs)->fs_nsect * (fs)->fs_trackskew + \
(bno) * NSPF(fs) % (fs)->fs_spc % (fs)->fs_nsect * (fs)->fs_interleave) % \
(fs)->fs_nsect * (fs)->fs_nrpos / (fs)->fs_npsect)
First I assume that the hard disk sectors are numbered from one cylinder
to another, moving from top track to the bottom track before moving to the
next cylinder. (This is the vertical mapping as described at the site
http:/www.tomshardware.com/hdd.html). Otherwise, the macro might not make
any sense.
So bno * NSPF % fs_spc / fs_nsect will give us the track number and bno *
NSPF % fs_spc % fs_nsect will give us the sector count within that track.
However, the rest part of the macro is confusing to me.
In particular, why we multiply fs_trackskew? What is the meaning of
fs_nrpos (number of rotational positions in a single track?) Why we
divide by fs_nsect after multiplying by fs_interleave? I assume that
fs_nsect + unused sectors = fs_npsect and the C operators %, /, and * all
have same precedence.
Any help is appreciated.
--------------------------------------------------
| Zhihui Zhang, http://cs.binghamton.edu/~zzhang |
| Dept. of Computer Science, SUNY at Binghamton |
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