(Foutrelis didn't see the "reply to all" button :) )
Begin forwarded message:
Dieter Plaetinck wrote:
As far as I understand: SI prefixes are, and always have been multiples of 10. ( eg 10^3,10^6)
But, many people have abused SI prefixes to denominate numbers like 2^10 (1024), 2^20 etc. Eg OS vendors and many software developers have used this definition, while hard disk vendors used the real meaning
So, along came IEEE to introduce prefixes for these binary multiples, so that we can use SI prefixes only for what they are meant (ieee at the same time also definined some more things like b=bit, B=byte, etc)
That's pretty much my understanding too.
But, if I create in the installer a partition of the size 10.000MiB:
echo $(((2**20)*10000)) 10485760000 <-- this is the amount of bytes I expect to see on the blockdevice.
fdisk -l says: 10487232000 bytes Why is it not the same?
Perhaps rounding takes place? 10487232000 bytes is about 10001.4 MiB, which is pretty close to the 10000 you specified.
Also a df -m (-m is not documented? but it says "1M" blocks) of the filesystem (ext2) yields 9845M. If 1M means 10^6 then this is wrong because it would be way too few, and if it means 1MiB than it's also still wrong. how much space can you loose by formatting a block device as ext2? Even after setting reserved blocks to 0 (tune2fs -r 0 /dev/sda3) It still is 9845M. (the amount available is 9823M
I believe there is around one percent file system overhead (I don't have any documentation to support my claim though). And probably this is used to store inodes (amongst other information?):
[root@archiso ~]# tune2fs -l /dev/sda3 | grep -P '(Inode count)|(Inode size)' Inode count: 640848 Inode size: 256 [root@archiso ~]# echo $((640848*256/2**20)) 156
So, with each inode taking up 256 bytes of space, and a total of 640848 inodes, it adds up to 156 MiB. Add that to what df -m reports (9845 MiB) and you get ~10000 MiB.
PS: If someone knows more about this stuff, please do share your knowledge! :3