Date: Mon, 1 Dec 1997 09:10:02 -0800 (PST) From: Bruce Evans <bde@zeta.org.au> To: freebsd-bugs Subject: Re: kern/5186: bug in diskslice_machdep.c (fix included) Message-ID: <199712011710.JAA21660@hub.freebsd.org>
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The following reply was made to PR kern/5186; it has been noted by GNATS.
From: Bruce Evans <bde@zeta.org.au>
To: andre.albsmeier@mchp.siemens.de, FreeBSD-gnats-submit@FreeBSD.ORG
Cc: Subject: Re: kern/5186: bug in diskslice_machdep.c (fix included)
Date: Tue, 2 Dec 1997 03:57:25 +1100
> Either the code or the comment above it is wrong in sys/i386/isa/diskslice_machdep.c:
They still look correct to me.
> /*
> * If ssector1 is on a cylinder >= 1024, then ssector can't be right.
> * Allow the C/H/S for it to be 1023/ntracks-1/nsectors, or correct
> * apart from the cylinder being reduced modulo 1024. Always allow
> * 1023/255/63.
> */
> if (ssector < ssector1
> && ((chs_ssect == nsectors && dp->dp_shd == ntracks - 1
> && chs_scyl == 1023)
> || (secpercyl != 0
> && (ssector1 - ssector) % (1024 * secpercyl) == 0))
> || (dp->dp_scyl == 255 && dp->dp_shd == 255
> && dp->dp_ssect == 255)) {
>
> Since we can't enter 255 for the number of sectors, I think it's the code.
dp_ssect consists of a 6-bit sector number and the top two bits of the head
number. dp->ssect == 255 means sector 63, head >= 768. In particular, it
is 255 for 1023/255/63 geometry.
Bruce
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