Re: [Info-mtools] Bug: mformat does not format FAT32 correctly

[Pali Rohár]

On Monday 04 December 2017 19:29:53 Pali Rohár wrote:
> Hi!
> 
> I observed 3 problems when I tried to create FAT32 disk image with
> mformat tool:
> 
> 
> 1) mformat is not able to format disk images without specifying C/H/S
> geometry. Nowadays there is no hard disk or SSD disk which use C/H/S
> geometry and therefore asking user for such thing does not make sense.
> 
> So what I used is to specify 64 heads and 32 sectors per track just to
> easily calculates number of tracks on disk. For disk with logical sector
> size as 512 bytes (which is probably for all HDD and SSD), number of
> tracks would be size of disk in megabytes (64*32*512 = 1 megabyte).
> 
> So mformat parameters: -h 64 -s 32 -t SIZE_IN_MB
> Probably it is also needed to specify (for hard disks): -m 0xf8
> 
> If you do not enter this C/H/S geometry mformat just show error message:
> 
> mformat: Unknown geometry (You must tell the complete geometry of the
> disk, either in /etc/mtools.conf or on the command line)
> 
> I think that except this error message, mformat for hard drives should
> use above C/H/S auto-calculation.
> 
> 
> 
> 2) mformat refuse to create FAT32 if disk size is less then 257 MB.
> 
> Test to reproduce (-F means to format as FAT32):
> 
>   $ truncate -s 100M fat.img
>   $ mformat -F -h 64 -s 32 -t 100 -m 0xf8 :: -i fat.img
> 
> mformay show just error message:
> 
> Too few clusters for this fat size. Please choose a 16-bit fat in your
> /etc/mtools.conf or .mtoolsrc file
> 
> Fix is to set size of cluster (in number of sectors) to 1. So mformat
> parameter: -c 1
> 
> It is pity that mformat cannot calculate cluster size for smaller FAT32
> disks (when size is less then 257 MB). Note that mformat can calculate
> it correctly for FAT16 (when -F is not used).
> 
> 
> 
> 3) Cluster size is not calculated correctly according to Microsoft FAT
> specification. This also answer why there is a bug 2).
> 
> Function which calculate cluster size is named calc_cluster_size and can
> be found in mformat.c source code. Relevant part for FAT32 is:
> 
>       switch(abs(fat_bits)) {
> ...
>               case 32:
>                       Fs->cluster_size = 8;
>                       /* According to
>                        * 
> http://support.microsoft.com/support/kb/articles/q154/9/97.asp
>                        * Micro$oft does not support FAT32 with less than 4K
>                        */
>                       return;
> ...
> 
> It means that cluster size is hardcoded to 8 sectors for FAT32. This is
> also reason for problem 3). For FAT16 there is autodetection code. I do
> not know what that comment mean or what it try to refer as link is dead.
> 
> So I would like to know, why is hardcoded 8 sectors as cluster size for
> FAT32? I suspect this is a bug which needs fixing. At least to have
> working mformat in FAT32 mode for small disk (or disk images).

Also please note that current versions of Windows (probably since 2000)
allow user to specify cluster size in Format dialog also for FAT32. And
default value depends on disk size, it is not hardcoded to 8 (=4kB).

See also:
https://support.microsoft.com/en-us/help/192322/description-of-default-cluster-sizes-for-fat32-file-system
https://support.microsoft.com/en-us/help/140365/default-cluster-size-for-ntfs--fat--and-exfat

So I would really suggest to change above hardcoded value 8 to either
mapping table or use similar algorithm as is already present for FAT16
in mformat.c

> Because, in Microsoft FAT32 specification (fatgen103.doc) on page 20 for
> FAT32 is written calculation table. For disk size there is cluster size
> (in number of sectors), see table below:
> 
> 32.5 MB - 260 MB   cluster_size =  1
>  260 MB -   8 GB   cluster_size =  8
>    8 GB -  16 GB   cluster_size = 16
>   16 GB -  32 GB   cluster_size = 32
>   32 GB -   2 TB   cluster_size = 64
> 
> Note that in that Microsoft FAT specification is written: "The rest of
> this section is totally specific to drives that have 512 bytes per
> sector.". So there is need to recalculate values when (logical) sector
> size is not 512 bytes.
> 

-- 
Pali Rohár
address@hidden