(C) Labutin D., Nizhny Novgorod
region, Dzerzhinsk
You have already written to the ZX-95 REVIEW N5
on page 53 that you can work with the regime interrupt IM 2
and at the same address to the disk.
This way I have not approached,
since, working in the mode of 128 kb,
interrupt handler with
addresses of 65,129 or 65,524 have not, because pages there
constantly changing, and the other
appropriate register values I
I have not picked it up. Then I thought
another way. All appeals to the
disk (CALL # 3D13, or similar)
is replaced by CALL DOS. Sub-DOS is as follows:
140.
DOS PUSH AF
LD A, 1
LD (FLAG), A
POP AF
CALL # 3D13
PUSH AF
XOR A
LD (FLAG), A
POP AF
RET
FLAG DEFB 0
A routine interrupts looks like this:
IM2 PUSH AF
LD A, (FLAG)
OR A
JR Z, IM21
POP AF
PUSH HL
LD HL, # 3D2F
EX (SP), HL
PUSH AF
PUSH BC
IM21 ............... ; Preservation registers
CALL PROG; treatment interruption
............... ; Restore registers
POP BC
POP AF
RET
2
Ed.: Good idea, only you do not
take into account the fact that the TR-DOS ROM is ROM
BASIC even when performing such simple
operations, such as reading groups of sectors. For
This routine is used at
# 5CC2, consisting of only one team
RET. Placed on the stack address of the subroutine, the
registers are loaded options, and done JP # 5CC2. Return of the
subroutine is at # 3D2F. To have your method to work, should be
at # 5CC2 put JP on the following subroutine: 140.
POP HL; take the address of the called subroutine
PUSH AF; keep AF
SUB A; working ROM BASIC
LD (FLAG), A
POP AF; recovery AF
PUSH HL; store address of the calling subroutine
LD HL, L1; put on a stack of new return address
; From subroutine
EX (SP), HL; and restore the address of the subroutine
PUSH HL; pushed onto the stack address of subroutine
LD HL, (23810); restore HL
RET; move on to the subroutine
L1 PUSH AF; keep AF
LD A, # 01; working ROM TR-DOS
LD (FLAG), A
POP AF; recovery AF
RET; return to the TR-DOS ROM
2
A few words about the restoration of HL.
Subroutine call ROM BASIC uses
variables 23810 and 23812 to save
HL and DE, respectively, therefore, for
call # 5CC2 in these variables will be copies of these
registers. This fact and used the above routine.
Corr.: I'm writing a graphic adventyuru (certainly not me
one, but I program), and
one point realized that the floppy disk in
640 KB enough for graphics.
Then I decided to switch to non-standard format. In one lane 5
sectors of 1024 bytes. Obtained disk size of 800 KB. I do not
remember what program I've formatted floppy disk giving the
sector number 1, 2, 3, 4, 5.
It seemed that the deed is done,
but there it was - the fifth sector
unreadable. You can do the following:
140.
LD E, 0, or 1,2,3
LD C, 5
LD B, 1
LD D, 5, or any
; Track
LD HL, 40000
CALL # 3D13
But if you try:
LD BC, # 0105
LD HL, 40000
LD DE, # 0504
CALL # 3D13
2
that sector is unreadable. Here I
remembered that in the IS DOS 5 sectors per track for the
kilobyte. After analyzing the contents of a track floppy disk,
formatted IS DOS, I was very surprised. There
sector have the numbers 1, 2, 3, 4,
9. Why 9, not 5? Reformatting the disk in this way,
I was no more similar
problems. I wrote the following
uploader:
140.
DOSLOAD PUSH BC
PUSH DE
PUSH HL
LD A, E
CP 4
JR NZ, DOSLD1
LD E, 8
DOSLD1 LD B, 1
CALL # 3D13
POP HL
INC H
INC H
INC H
INC H
POP DE
INC E
LD A, E
CP 5
JR NZ, DOSLD2
INC D
LD E, 0
DOSLD2 POP BC
DJNZ DOSLOAD
LD (# 5CF4), DE
RET
2
Before the call: HL = address
loading, B = number of sectors
(1024 bytes), C = 5 - reading,
6 - entry, D = path, E =
sector (from 0 to 4). But downloading
walking very slowly, and I've placed the sector on the track in
the following order: 1, 4, 2, 9, 3, and rate increased by 2
times. Maybe this information will prove useful to somebody,
and someone can explain why need to use nine sectors, rather
than five.
Prim Ed.: Strange story. Ninth
sector, or the fifth, but at least the 130th, all
should work. Perhaps the lack of Dmitri in the format program.
Dmitry, try to format the drive so how do you like the same
program, and then look at it with the function Analyse
Track ADS programs or RDS. Likely
You'll find the answer to your question. Now on
why in the IS-DOS used in the ninth
sector. It is necessary for the implementation of
Startup of the system. The principle is. Before any operation
with TR-DOS disk loads nine sectors zero track in a specially
allocated for this buffer at # 5D25. The buffer has a length of
257 bytes. Why not 256 - can only say the creators of TR-DOS.
During this buffer area will be based on descriptors channels.
TR-DOS does not can read part of the sector, it loads
the whole thing, no matter how long it was. In
result of sector length of more than 256
byte overwrites a descriptor channels.
But the TR-DOS does not know and tries to
identify the disc. The first thing she
checks in the sector of bytes at offset
231, and if it does not equal 16, it is
immediate interruption of operation with the issuance of
Posts Disc Error. Issue Post
goes through the channel K. Here's the startup mechanism is
triggered. Enough to cell offset # 0101 and # 0102 to record
address of your boot, which is located in the same sector. The
loader will start, and there really do whatever you want.
*
