I need to Know the What is the Maximum volume size of VFAT (Maximu file system size to support VFAT).
I know the Maximum file size of VFAT is 2GB, I Need the Maximum Volume size of VFAT.
Rgards,
Vamsi.
2TB for 512 byte sector size and 16TB for a 4K sector size.
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So as far as i know, MBR size is 512 bytes and it ends with 0x55AA.
My questions are :
How common is it for disks to have a sector size bigger than 512 bytes ? (eg disks used in servers, etc)
How will MBR get stored if sector size is for example 4096? is the 4095th byte equal to 0x55 or the 511th byte?
Do i need to check for sector size before reading MBR in my C code? is this something i need to worry about, or just reading 512 first bytes of a disk is enough ?
Most of the devices even if the underlying hardware uses longer blocks provide "standard" interface where block size 512 bytes.
If the size of the sector is larger than 512 bytes the MBR will look exactly the same bytes. All data will be at the same offsets. Simply the remaining part of the sector will not be used.
If the sector size is smaller then 512 bytes then MBR will occupy more than one sector.
Do i need to check for sector size before reading MBR in my C code?
Very often the single sector is minimum chunk you need to read, so it is good to know its size as you need to allocate a buffer to store this data.
Is there a way to identify the type of a FAT partition (if it is 16 or 32) only by reading its boot sector?
thanks.
Not by reading the boot sector - You need to look into the file system itself.
Find the number of clusters. The file system subtype can be determined by this number:
less than 4086: FAT12
equal or more than 4086: FAT16
more than 65525: FAT32
If the sectors per FAT word in the FAT12/FAT16 BPB is zero, it is FAT32. (Regardless of the actual FAT size, FAT32 uses the EBPB's sectors per FAT dword.) Likewise, if the number of root directory entries word is zero, it is FAT32.
ll /srv/node/dcodxx/test.sh
-rw-r--r--. 1 root root 7 Nov 5 11:18 /srv/node/dcodxx/test.sh
The size of the file is shown in bytes. This file is stored in an xfs filesystem with block size 4096 bytes.
xfs_info /srv/node/sdaxx/
meta-data=/dev/sda isize=256 agcount=32, agsize=7630958 blks
= sectsz=4096 attr=2, projid32bit=0
data = bsize=4096 blocks=244190646, imaxpct=25
= sunit=0 swidth=0 blks
naming =version 2 bsize=4096 ascii-ci=0
log =internal bsize=4096 blocks=119233, version=2
= sectsz=4096 sunit=1 blks, lazy-count=1
realtime =none extsz=4096 blocks=0, rtextents=0
Does this mean that a block can house more than one file, if not what happens to the remaining bytes (4096-7)?
Also, where is the 256 bytes reserved for an inode stored, if it stored in the same block as the file, shouldn't the file size be larger(256+7)?
File data is stored in units of the filesystem block size, and no block sharing is currently possible across multiple files on XFS. So used disk space is always the number of bytes in the file rounded up to the next block size - a 1-byte file will consume 4k of diskspace on a 4k block size filesystem.
The inode itself contains file metadata such as size, timestamps, extent data, etc - and on xfs it can also contain extended attribute information.
The on-disk inode is separate from the file data blocks, and will always consume 256 bytes on a filesystem with 256 byte inodes, regardless of the amount of metadata used. If more than 256 bytes is required to store additional extent information or extended attribute data, additional filesystem-block-sized metadata blocks will be allocated.
Does this mean that a block can house more than one file, if not what happens to the remaining bytes (4096-7)?
A block cannot contain more than one file. If a file is bigger than one block, multiple blocks are used.
Modern filesystems like XFS have a functionality called "inline", where files small enough (no more than 60 bytes) can be stored in the inode, in the space taken to store pointers to the blocks.
where is the 256 bytes reserved for an inode stored, if it stored in the same block as the file, shouldn't the file size be larger(256+7)?
Inode information is stored in the inode table.
i know the cluster size of a USB flash drive can be changed , can we change the sector size too ??
Sector size isn't a configurable parameter in ATA/SATA/SCSI/etc devices and, from my experience, USB flash drives implement one of these protocols. The sector size is reported by the device itself but, even if you could set it to something other than 512, you would likely run into a latent bug somewhere in a driver or file system package that assumed a sector size of 512.
There are real reasons for using a sector size like 512, for example, addressing of larger sectors can be done more quickly and efficiently (not just in time but in size/space as well). Throughput to these devices is also better with something like 512. Consider that, if you could set sector size to something like 16-bytes, you might have less wasted space with 16-byte sectors compared to having a number of half-full 512-byte sectors, but your throughput to the device would probably be worse. In fact, writing a single 16-byte sector would only be slightly faster than writing a single 512-byte sector. On the other hand, writing 32 16-byte sectors (to write a total of 512 bytes) would likely take longer than writing a single 512 byte sector simply due to the overhead associated with transferring multiple sectors.
I would suggest you buy a larger USB flash drive if you are worried about wasting space with 512-byte sectors.
In reading about the Unix FFS, I've read that 10% of the disk space is reserved so that files' data blocks can be ensured to be in the same cylinder group. Is this still true with filesystems like ext2/ext3, is there space reserved so that files' data blocks can all be in the same block group? Is it also 10%? or does it vary? Also, is the same true for journaling filesystems as well? Thank you.
first of all i think that ext filesystems implement the same notion of a cylinder group,
they just call it block group.
to find out about it , you can fdisk the partition to find your actual block count
and blocks/group number .Then the number of block groups = block count / (block/group).
They are used in exactly the same way as FFS cgs (to speed up access times).
Now journaling IMO has nothing to do with this operation, except that it actually wastes
some more space on your disk :). As far as i understand , soft updates which is the BSD solution to the problem that a journal would solve in typical ext filesystems, don't require extra space , but are tremendously complex to implement and add new features on (like resizing).
interesting read:
ext3 overhead disclosed part 1
cheers!
My data for fresh ext2 images are:
Size Block size Bl/Gr Total bytes Free bytes Ratio
1MB 1024 8192 1048576 1009664 0.03710
10MB 1024 8192 10485760 10054656 0.04111
100MB 1024 8192 104857600 99942400 0.04688
512M 4096 32768 536870912 528019456 0.01649
1G 4096 32768 1073741824 1055543296 0.01695
10G 4096 32768 10737418240 10545336320 0.01789
So, it's quite predictable that the space efficiency of an Ext2 filesystem depends on block size due to layout described in the above answer: filesystem is a set of block groups, for each group its size is determined as count of blocks which can be described by a 1-block bitmap => for a 4096 byte block there are 8 * 4096 blocks.
Conclusion: for ext2/ext3 family of filesystems average default consumption of space depends on block size:
~ 1.6 - 1.8 % for 4096 byte blocks, ~ 4 % for 1024 ones