From: Andries.Brouwer@cwi.nl
To: aia21@cam.ac.uk
Cc: linux-kernel@vger.kernel.org
Subject: Re: Getting the number of 512-byte sectors?
Date: Mon, 22 Jan 2001 02:31:04 +0100 (MET) [thread overview]
Message-ID: <UTC200101220131.CAA176461.aeb@vlet.cwi.nl> (raw)
From: Anton Altaparmakov <aia21@cam.ac.uk>
Subject: Getting the number of 512-byte sectors?
My question is how to get the _real_ number of sectors of a partition from
within a file system. I.e. we are starting only with the knowledge of the
struct super_block for the partition.
As you conjectured already, you find this info in the gendisk->part[] array,
in the start_sect and nr_sects fields.
Locate the correct gendisk structure by performing a walk of gendisk_head
as done in drivers/block/blkpg.c::get_gendisk(kdev_t), (Could we make this
function to not be private to blkpg.c?), then get the hd_struct for the
partition and obtain the values from that. In code:
struct gendisk *g = get_gendisk(my_super_block->s_dev)
struct hd_struct *hd = &g->part[MINOR(my_super_block->s_dev)];
wanted "subtracted" values:
hd->start_sect
hd->nr_sects
Am I missing something?
No, I agree completely.
Note: I know I can normally get the number of sectors from the boot sector
of the partition
Not necessarily. The partition table need not be of DOS type.
Or, even if it is, the kernel's tables may be set up dynamically.
> Could we make this function to not be private to blkpg.c?
I think this area will change rather drastically in 2.5.*.
Andries
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next reply other threads:[~2001-01-22 1:31 UTC|newest]
Thread overview: 3+ messages / expand[flat|nested] mbox.gz Atom feed top
2001-01-22 1:31 Andries.Brouwer [this message]
-- strict thread matches above, loose matches on Subject: below --
2001-01-21 17:05 Anton Altaparmakov
[not found] <20001002032923.B17146@veritas.com>
[not found] ` <00100202491203.00767@gimli>
[not found] ` <news2mail-39D7A4E9.26E4DA3C@innominate.de>
[not found] ` <00100201582602.00767@gimli>
[not found] ` <20001002022728.A17106@veritas.com>
2001-01-21 16:16 ` Anton Altaparmakov
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