From: Ryan Wang Subject: How to understand super.c/descriptor_loc for ext2/ext3/ext4? Date: Thu, 7 Jun 2012 21:37:07 +0800 Message-ID: Mime-Version: 1.0 Content-Type: multipart/mixed; boundary="===============4225155415163879039==" To: linux-ext4@vger.kernel.org, kernelnewbies@kernelnewbies.org Return-path: List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , Sender: kernelnewbies-bounces@kernelnewbies.org Errors-To: kernelnewbies-bounces@kernelnewbies.org List-Id: linux-ext4.vger.kernel.org --===============4225155415163879039== Content-Type: multipart/alternative; boundary=90e6ba6153dad7b4bf04c1e1fa59 --90e6ba6153dad7b4bf04c1e1fa59 Content-Type: text/plain; charset=ISO-8859-1 Hi, I'm reading the source code for ext2/ext3/ext4, and cannot figure out how descriptor_loc() runs: 1600 static ext3_fsblk_t descriptor_loc (struct super_block *sb ,1601 ext3_fsblk_t logic_sb_block ,1602 int nr )1603 {1604 struct ext3_sb_info *sbi = EXT3_SB (sb );1605 unsigned long bg , first_meta_bg ;1606 int has_super = 0;1607 1608 first_meta_bg = le32_to_cpu (sbi ->s_es ->s_first_meta_bg ); 1) Who can setup the field ->s_first_meta_bg, mke2fs? And what does it mean? What's a meta block group? 1609 1610 if (!EXT3_HAS_INCOMPAT_FEATURE (sb , EXT3_FEATURE_INCOMPAT_META_BG ) ||1611 nr < first_meta_bg ) 2) In which case, we can have "nr < first_meta_bg"? And how does EXT3_FEATURE_INCOMPAT_META_BG affect the physical disk layout? 1612 return (logic_sb_block + nr + 1);1613 bg = sbi ->s_desc_per_block * nr ;1614 if (ext3_bg_has_super (sb , bg ))1615 has_super = 1;1616 return (has_super + ext3_group_first_block_no (sb , bg )); 3) It always return the first block no of the specific bg, right? Then its caller ext3_fill_super will read in the disk block. It seems that ext3_fill_super just read many copies of one same block (1st block), right? I'm sure I'm wrong with it, but cannot find out the right way. 1617 } Any explanation are welcome. thanks, --90e6ba6153dad7b4bf04c1e1fa59 Content-Type: text/html; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable Hi,

I'm reading the source code for ext2/ext3/ext4, and cannot f= igure
out how descriptor_loc() runs:

1600static ext3_fsblk_t desc=
riptor_loc(struct super_block *sb,
1601                                    <=
a href=3D"http://lxr.linux.no/linux+*/+code=3Dext3_fsblk_t" class=3D"sref">=
ext3_fsblk_t logic_sb_block,
1602                                    i=
nt nr=
)
1603{
1604        struct ext3_sb_info *sbi =3D=
 EXT=
3_SB(=
sb);
1605        unsigned long bg, first_meta_bg<=
/a>;
1606        int has_super =3D 0;
1607
1608        first_meta_bg =3D le32_=
to_cpu(sbi->s_es->s_first_meta_bg); 


1) Who can setup the field ->s_first_meta_bg, mke2fs? And what does =
it mean?
=A0=A0 What's a meta block group? 

1609
1610        if (!EXT3_HAS_I=
NCOMPAT_FEATURE(sb, EXT3_FEATURE_INCOMPAT_META_BG) ||
1611            nr < first_meta_bg)

2) In which case, we can have "nr < first_meta_bg"?
=A0=
=A0 And how does EXT3_FEATURE_INCOMPAT_META_BG affect the physical disk lay=
out?


1612                retur=
n (logic_sb_block + nr + 1);
1613        bg =3D sbi->s_desc_per_block * nr;
1614        if (ext3_bg_has_super(sb=
, bg)=
)
1615                has_super =3D 1;
1616        return (has_super + ext3_group_first_block_no(sb, bg));

3) It always return the first block no of the specific bg, right?
=
=A0=A0 Then its caller ext3_fill_super will read in the disk block.
=A0=
=A0 It seems that ext3_fill_super just read many copies of one same block=A0=A0 (1st block), right?

=A0=A0 I'm sure I'm wrong with it, but cannot find out the right wa=
y.

1617}

Any explanation are welcome.

thanks,
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