(C) M. Kuzmin, Moscow Region.
REVIEW In an article about Roshchina
reduction in the time format. The above is there a way
theoretically good, the practice worse: on different drives
while formatting a single track different, and therefore bind
to the time zero can not be done. But in general, as can be
done during the interruption formatting, if the program
will hang in DOS'e on the block
record??
Writing sectors simultaneously
with formatting - just a waste of time. It lost to scan sectors
to format (To detect # F5, # F6, # F7).
Yes, of course, if these sectors do not have to track, then the
payoff will be, but if any one sector "Unclean" (especially
from the past on the track), it still lost the turnover drive
on his record - loss is clearly greater.
Ca. Ed.: Indeed, if the formatting a track by means of
TR-DOS, you can not interrupt it. Hence, to directly control
the VG93. An example of an algorithm format:
- Written in the instruction register code to VG93
Manda formatting;
- Grant to SH bytes corresponding
selected format, as long as
will be formatted in the last sector;
- Use the force
break;
As can be seen, while formatting a
track does not play a role, because it is generally not used.
But it turns out one unpleasant fact: not enough speed Z80 in
order to give on SH bytes to format. Indeed, the disc rotates
at a speed of 300 rpm for 1 turnover is spent 0.2 seconds. or
700,000 cycles Z80. Since the track is recorded about 7000
bytes to write each of them must be spent not
more than 100 cycles Z80. Meanwhile, the fragment
program that writes to the disk a
bytes will be performed at least 120 cycles:
140.
; Let the registers have the following meanings:
; C - # 7F (register address data)
; HL - # 20AF (address n / n write bytes)
; IX - address the following command to which
, It hands over control after burning
; Bytes.
LD D, B 3, 7 clocks
PUSH IX 3, 1915
PUSH HL; 3 11
JP # 3D30; 3 10
# 3D30: RET; 3 10
# 20AF: LD B, March 1, 7
# 20B1: IN A, (255); 3 11
AND # C0; 3 7
JR Z, # 20B1; 3 7 / 12
RET M 3, 5 / 11
OUT (C), D 3, 1912
DJNZ # 20B1; 3 8
RET; 3 10
2
And all because it registers the SH appear in the address
space only at the moment of TR-DOS. If there was free access to
the registers, without the involvement of TR-DOS, then
ultra-fast format would be a reality!
All of the above applies to the record
sectors simultaneously with formatting.
The check for the presence of the sector of bytes # F5, # F6 #
F7 and should be done direct formatting, so
that the extra time it takes. Yes, if these bytes will be
found, lost another turnover disc for recording sector. But not
so often they occur - for example, on a disk with text files of
their may be absent.
Corr.: Now, about rebuilding the screen for one interrupt:
as mentioned in various sources, it is impossible, but in
specific cases, it is a very really. Take, for example, ACTION,
LIFE SUX - demo. There's no just rebuilt the whole screen, but
in a grid is displayed under a rather large sprite in the
second line runs. Done this is simple: how could
note that these screens are composed of
identical segments (for example,
in LIFE SUX 2 * 2 familiarity). Program to output a single line
is this:
LD SP, scradr +1
LD HL, BYTE
PUSH HL; 16 times (16 * 2 = 32)
As can be seen, the output of one line
takes 11 * 16 +10 +10 = 191 bar, and
as one line of video controller passes for 224 cycles,
there are still plenty of time (if
take into account also the upper and BORDER
below).
HELP ME! I have done recently
scroller (a hobby I have is:
do different VIEWER'y, text scrollers, help'y, etc.), and
faced with a problem:
stack was repositioned on the screen
(Read / write lines), but were
included IM 2, and although the program
Interrupt rearranged in the stack buffer, the screen is still
spoiled, because Z80 when you call IM 2 still puts on a stack
of two bytes ... In the ZX-FORMAT'e Mednonogov said that the
combined output Music and sprites through the stack -
can anyone help
this issue - would be very grateful ...
Ca. Ed.: If the transfer of data through the stack and
allowed termination, you must carefully plan time distribution
Z80. Suppose that the interval between two interruptions 70,000
cycles procedure interrupt handling is
10,000 cycles, and the procedure for SCR, with
stack shear screen content, is 100,000 cycles. Then it is
better to break this procedure into several parts, say,
5 for 20000 cycles each, and the program
will be as follows:
HALT; performed procedure for handling
; Interrupt - 10000 cycles
CALL SCR1; 20000
CALL SCR2; 20000
CALL SCR3; 20000
HALT; 10000
CALL SCR4; 20000
CALL SCR5; 20000
CALL SCR1; 20000 (next shift)
...
