I am now programming on a 8 bits Z80 computer with CP/M 2.2, (as a hobby) and the floppy disk format is IBM, 40 tracks, 8 sectors per track, 512 bytes per sector. free space is 154 Ko on each face of the disk.
Why the sectors are indexed 1 to 8 (and not zero to seven, as usually is seen with computers)?
The catalog of the floppy disk is on the track 1 (sector 1 to 4, 64 entries).
I'm wondering is the catalog on track zero?
Is the zero track reserved to included a system (as track 0 & 1 are reserved to the system on a CP/M floppy disk, and catalog is on track 2)?
I asked this because for example, on APPLE II, sectors start at zero (0 to F).
The computer I use is an AMSTRAD CPC 6128 (1985). It is a 8 bits with Zilog Z-80, with CP/M OS 2.2 and 3.0 available.
The floppy disk format is pretty rare : 3 inches. (not 3 1/2).
3 FORMATS are available : -CP/M : 40 tracks (0 to 39) 9 sectors per track (named &41 to &49) 169 Ko on each face. -DATA : 40 tracks (0 to 39) 9 sectors per track (named &C1 to &C9) 178 Ko on each face. -IBM : 40 tracks (0 to 39) 8 sectors per track (named 1 to 8) 154 Ko free space on each face.
A 3 inch floppy disk is différent of a 3 1/2 floppy disk. To read the side B: I have to eject the diskette and turn it myself then put it again in the disk drive! but it works very well.
The tracks start at 0 to 39. I am just curious to know why sectors start at 1, and why the track 0 is not use for the directory on the IBM format...
For example, CP/M format is very simple : directory on track 2, system CP/M on track 0 and 1.
I am an experienced assembly programmer, and I think it would be logic that sectors start at zero, as tracks do. The AMSTRAD CPC computer can accept a 5 1/4 disk drive too. I bought one yesterday. But tracks,sectors and formats will be the same of course on 5 1/4 disquettes.
p.s: sorry for my english, it is not my native language. ^^